{"id":126033,"date":"2023-09-12T06:21:00","date_gmt":"2023-09-12T00:51:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126033"},"modified":"2023-12-01T10:34:36","modified_gmt":"2023-12-01T05:04:36","slug":"rd-sharma-class-10-solutions-chapter-5-exercise-5-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-exercise-5-1\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126042\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-5-Exercise-5.1.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-5-Exercise-5.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-5-Exercise-5.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1:&nbsp;<\/strong>This exercise includes problems that require you to determine all of the trigonometric ratios when only one is given. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> has all of the solutions to this and other chapters. Students can get <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-trigonometric-ratios\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 5 Trigonometric Ratios<\/strong><\/a> Exercise 5.1 PDF if they have any questions about the solution or concept.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d059d215d12\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-exercise-5-1\/#where-can-i-download-rd-sharma-class-10-maths-solutions-chapter-5-exercise-51-free-pdf\" title=\"Where can I download RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.1 free PDF?\">Where can I download RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.1 free PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-exercise-5-1\/#is-it-important-to-study-all-of-the-concepts-included-in-rd-sharma-class-10-solutions-chapter-5-exercise-51\" title=\"Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1?\">Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-5-exercise-51-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-5-Exercise-5.1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-5-Exercise-5.1.pdf\">RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-5-exercise-51-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.1- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. In each of the following, one of the six trigonometric ratios s given. Find the values of the other trigonometric ratios.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 1\" width=\"192\" height=\"172\"><\/strong><\/p>\n<p><strong>(i)&nbsp;sin A = 2\/3<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have,<\/p>\n<p>sin A = 2\/3 \u2026\u2026..\u2026.. (1)<\/p>\n<p>As we know, by sin definition,<\/p>\n<p>sin A&nbsp;=&nbsp;&nbsp;Perpendicular\/ Hypotenuse&nbsp;= 2\/3&nbsp;\u2026.(2)<\/p>\n<p>By comparing eq. (1) and (2), we&nbsp;have<\/p>\n<p>Opposite side = 2 and Hypotenuse = 3<\/p>\n<p>Now, on using Pythagoras theorem in \u0394&nbsp;ABC<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2 +<\/sup>&nbsp;BC<sup>2<\/sup><\/p>\n<p>Putting the values of perpendicular side (BC) and hypotenuse (AC) and for the base side as (AB), we get<\/p>\n<p>\u21d2 3<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ 2<sup>2<\/sup><\/p>\n<p>AB<sup>2&nbsp;<\/sup>= 3<sup>2<\/sup>&nbsp;\u2013 2<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 9 \u2013 4<\/p>\n<p>AB<sup>2 =&nbsp;<\/sup>5<\/p>\n<p>AB = \u221a5<\/p>\n<p>Hence, Base =&nbsp;\u221a5<\/p>\n<p>By definition,<\/p>\n<p>cos A = Base\/Hypotenuse<\/p>\n<p>\u21d2 cos A = \u221a5\/3<\/p>\n<p>Since, cosec A =&nbsp;1\/sin A = Hypotenuse\/Perpendicular<\/p>\n<p>\u21d2 cosec A = 3\/2<\/p>\n<p>And, sec A =&nbsp;Hypotenuse\/Base<\/p>\n<p>\u21d2 sec A =&nbsp;3\/\u221a5<\/p>\n<p>And, tan A =&nbsp;Perpendicular\/Base<\/p>\n<p>\u21d2 tan A =&nbsp;&nbsp;2\/\u221a5<\/p>\n<p>And, cot A =&nbsp;1\/ tan A = Base\/Perpendicular<\/p>\n<p>\u21d2 cot A =&nbsp;\u221a5\/2<\/p>\n<p><strong>(ii) cos A = 4\/5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have,<\/p>\n<p>cos A = 4\/5&nbsp;\u2026\u2026.\u2026. (1)<\/p>\n<p>As we know, by cos definition,<\/p>\n<p>cos A = Base\/Hypotenuse&nbsp;\u2026. (2)<\/p>\n<p>By comparing eq. (1) and (2), we&nbsp;get<\/p>\n<p>Base = 4 and Hypotenuse = 5<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2&nbsp;<\/sup>+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of base (AB) and hypotenuse (AC) and for the perpendicular (BC), we get<\/p>\n<p>5<sup>2<\/sup>&nbsp;= 4<sup>2&nbsp;<\/sup>+ BC<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 5<sup>2<\/sup>&nbsp;\u2013 4<sup>2<\/sup><\/p>\n<p>BC<sup>2&nbsp;<\/sup>= 25 \u2013 16<\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 9<\/p>\n<p>BC= 3<\/p>\n<p>Hence, Perpendicular = 3<\/p>\n<p>By definition,<\/p>\n<p>sin A&nbsp;=&nbsp;Perpendicular\/Hypotenuse<\/p>\n<p>\u21d2 sin A = 3\/5<\/p>\n<p>Then, cosec A =&nbsp;1\/sin A<\/p>\n<p>\u21d2 cosec A= 1\/ (3\/5) = 5\/3 = Hypotenuse\/Perpendicular<\/p>\n<p>And, sec A = 1\/cos A<\/p>\n<p>\u21d2 sec A =Hypotenuse\/Base<\/p>\n<p>sec A =&nbsp;5\/4<\/p>\n<p>And, tan A =&nbsp;Perpendicular\/Base<\/p>\n<p>\u21d2 tan A = 3\/4<\/p>\n<p>Next, cot A =&nbsp;1\/tan A = Base\/Perpendicular<\/p>\n<p>\u2234 cot A =&nbsp;4\/3<\/p>\n<p><strong>(iii) tan \u03b8 = 11\/1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, tan&nbsp;\u03b8 = 11\u2026..\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>tan&nbsp;\u03b8 = Perpendicular\/ Base\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we&nbsp;get;<\/p>\n<p>Base = 1 and&nbsp;Perpendicular = 5<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of base (AB) and perpendicular (BC) to get hypotenuse(AC), we get<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 1<sup>2<\/sup>&nbsp;+ 11<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 1 + 121<\/p>\n<p>AC<sup>2<\/sup>= 122<\/p>\n<p>AC= \u221a122<\/p>\n<p>Hence, hypotenuse = \u221a122<\/p>\n<p>By definition,<\/p>\n<p>sin = Perpendicular\/Hypotenuse<\/p>\n<p>\u21d2 sin&nbsp;\u03b8 = 11\/\u221a122<\/p>\n<p>And, cosec&nbsp;\u03b8&nbsp;= 1\/sin&nbsp;\u03b8<\/p>\n<p>\u21d2 cosec&nbsp;\u03b8&nbsp;= \u221a122\/11<\/p>\n<p>Next, cos&nbsp;\u03b8 = Base\/ Hypotenuse<\/p>\n<p>\u21d2 cos&nbsp;\u03b8 = 1\/\u221a122<\/p>\n<p>And, sec&nbsp;\u03b8 = 1\/cos&nbsp;\u03b8<\/p>\n<p>\u21d2 sec&nbsp;\u03b8 = \u221a122\/1 =&nbsp;\u221a122<\/p>\n<p>And, cot&nbsp;\u03b8 &nbsp;= 1\/tan&nbsp;\u03b8<\/p>\n<p>\u2234&nbsp;cot&nbsp;\u03b8 = 1\/11<\/p>\n<p><strong>(iv) sin \u03b8 = 11\/15<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, &nbsp;sin \u03b8 = 11\/15&nbsp;\u2026\u2026\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>sin \u03b8 = Perpendicular\/ Hypotenuse&nbsp;\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we get,<\/p>\n<p>Perpendicular = 11 and Hypotenuse= 15<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base (AB), we have<\/p>\n<p>15<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+11<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 15<sup>2&nbsp;<\/sup>\u2013 11<sup>2<\/sup><\/p>\n<p>AB<sup>2&nbsp;<\/sup>= 225 \u2013 121<\/p>\n<p>AB<sup>2&nbsp;<\/sup>= 104<\/p>\n<p>AB =&nbsp;\u221a104<\/p>\n<p>AB=&nbsp;\u221a (2\u00d72\u00d72\u00d713)<\/p>\n<p>AB= 2\u221a(2\u00d713)<\/p>\n<p>AB= 2\u221a26<\/p>\n<p>Hence, Base = 2\u221a26<\/p>\n<p>By definition,<\/p>\n<p>cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos\u03b8 = 2\u221a26\/ 15<\/p>\n<p>And, cosec&nbsp;\u03b8&nbsp;= 1\/sin&nbsp;\u03b8<\/p>\n<p>\u2234 cosec&nbsp;\u03b8&nbsp;= 15\/11<\/p>\n<p>And, sec\u03b8&nbsp;= Hypotenuse\/Base<\/p>\n<p>\u2234 sec\u03b8 =15\/ 2\u221a26<\/p>\n<p>And,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234&nbsp;tan\u03b8 =11\/ 2\u221a26<\/p>\n<p>And,&nbsp; cot \u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234&nbsp;cot\u03b8 =2\u221a26\/ 11<\/p>\n<p><strong>&nbsp;(v) tan \u03b1 = 5\/12<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, &nbsp;tan&nbsp;\u03b1 = 5\/12&nbsp;\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>tan \u03b1 = Perpendicular\/Base\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we&nbsp;get<\/p>\n<p>Base = 12 and Perpendicular side = 5<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of base (AB) and the perpendicular (BC) to get hypotenuse (AC), we have<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 12<sup>2<\/sup>&nbsp;+ 5<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 144 + 25<\/p>\n<p>AC<sup>2<\/sup>= 169<\/p>\n<p>AC = 13 [After taking sq root on both sides]<\/p>\n<p>Hence, Hypotenuse = 13<\/p>\n<p>By definition,<\/p>\n<p>sin&nbsp;\u03b1&nbsp;&nbsp;= Perpendicular\/Hypotenuse<\/p>\n<p>\u2234&nbsp;sin&nbsp;\u03b1 = 5\/13<\/p>\n<p>And, cosec&nbsp;\u03b1&nbsp;&nbsp;= Hypotenuse\/Perpendicular<\/p>\n<p>\u2234 cosec&nbsp;\u03b1&nbsp;= 13\/5<\/p>\n<p>And,&nbsp; cos&nbsp;\u03b1 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos&nbsp;\u03b1 = 12\/13<\/p>\n<p>And,&nbsp; sec&nbsp;\u03b1 =1\/cos&nbsp;\u03b1<\/p>\n<p>\u2234 sec&nbsp;\u03b1 = 13\/12<\/p>\n<p>And, tan&nbsp;\u03b1 = sin&nbsp;\u03b1\/cos&nbsp;\u03b1<\/p>\n<p>\u2234 tan&nbsp;\u03b1=5\/12<\/p>\n<p>Since, cot&nbsp;\u03b1 = 1\/tan&nbsp;\u03b1<\/p>\n<p>\u2234&nbsp;cot&nbsp;\u03b1 =12\/5<\/p>\n<p>&nbsp;<\/p>\n<p><strong>&nbsp;(vi) sin \u03b8 = \u221a3\/2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have,&nbsp;sin&nbsp;\u03b8 =&nbsp;\u221a3\/2&nbsp;\u2026\u2026\u2026\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>sin&nbsp;\u03b8 = Perpendicular\/ Hypotenuse\u2026.(2)<\/p>\n<p>On Comparing eq. (1) and (2), we get<\/p>\n<p>Perpendicular =&nbsp;\u221a3 and&nbsp;Hypotenuse = 2<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of perpendicular (BC) and hypotenuse (AC) and get the base (AB), we get<\/p>\n<p>2<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ (\u221a3)<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 2<sup>2<\/sup>&nbsp;\u2013 (\u221a3)<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 4 \u2013 3<\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 1<\/p>\n<p>AB = 1<\/p>\n<p>Thus, Base = 1<\/p>\n<p>By definition,<\/p>\n<p>cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos \u03b8 = 1\/2<\/p>\n<p>And, cosec&nbsp;\u03b8&nbsp;= 1\/sin&nbsp;\u03b8<\/p>\n<p>Or cosec&nbsp;\u03b8= Hypotenuse\/Perpendicualar<\/p>\n<p>\u2234 cosec&nbsp;\u03b8&nbsp;=2\/\u221a3<\/p>\n<p>And,&nbsp; sec&nbsp;\u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234&nbsp;sec \u03b8 = 2\/1<\/p>\n<p>And,&nbsp; tan&nbsp;\u03b8 = Perpendicula\/Base<\/p>\n<p>\u2234&nbsp;tan&nbsp;\u03b8 = \u221a3\/1<\/p>\n<p>And,&nbsp; cot&nbsp;\u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234&nbsp;cot&nbsp;\u03b8 = 1\/\u221a3<\/p>\n<p><strong>(vii) cos \u03b8 = 7\/25<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have,&nbsp;cos \u03b8 = 7\/25&nbsp;\u2026\u2026\u2026.. (1)<\/p>\n<p>By definition,<\/p>\n<p>cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>On Comparing eq. (1) and (2), we&nbsp;get;<\/p>\n<p>Base = 7 and Hypotenuse = 25<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of base (AB) and hypotenuse (AC) to get the perpendicular (BC),<\/p>\n<p>25<sup>2<\/sup>&nbsp;= 7<sup>2&nbsp;<\/sup>+BC<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 25<sup>2<\/sup>&nbsp;\u2013 7<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 625 \u2013 49<\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 576<\/p>\n<p>BC=&nbsp;\u221a576<\/p>\n<p>BC= 24<\/p>\n<p>Hence, Perpendicular side = 24<\/p>\n<p>By definition,<\/p>\n<p>sin \u03b8 = perpendicular\/Hypotenuse<\/p>\n<p>\u2234 sin \u03b8&nbsp;= 24\/25<\/p>\n<p>Since, cosec&nbsp;\u03b8&nbsp;= 1\/sin&nbsp;\u03b8<\/p>\n<p>Also, cosec&nbsp;\u03b8= Hypotenuse\/Perpendicualar<\/p>\n<p>\u2234 cosec&nbsp;\u03b8&nbsp;= 25\/24<\/p>\n<p>Since,&nbsp; sec&nbsp;\u03b8 = 1\/cosec&nbsp;\u03b8<\/p>\n<p>Also,&nbsp; sec \u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234 sec&nbsp;\u03b8 = 25\/7<\/p>\n<p>Since,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234&nbsp;tan&nbsp;\u03b8 = 24\/7<\/p>\n<p>Now,&nbsp; cot = 1\/tan&nbsp;\u03b8<\/p>\n<p>So,&nbsp; cot \u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234 cot&nbsp;\u03b8 = 7\/24<\/p>\n<p><strong>(viii) tan \u03b8 = 8\/15<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have,&nbsp;tan&nbsp;\u03b8 = 8\/15&nbsp;\u2026\u2026\u2026\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>tan \u03b8 = Perpendicular\/Base&nbsp;\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we get<\/p>\n<p>Base = 15 and Perpendicular = 8<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= 15<sup>2<\/sup>&nbsp;+ 8<sup>2<\/sup><\/p>\n<p>AC<sup>2&nbsp;<\/sup>= 225 + 64<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 289<\/p>\n<p>AC =&nbsp;\u221a289<\/p>\n<p>AC = 17<\/p>\n<p>Hence, Hypotenuse = 17<\/p>\n<p>By definition,<\/p>\n<p>Since,&nbsp; sin \u03b8 = perpendicular\/Hypotenuse<\/p>\n<p>\u2234&nbsp;sin \u03b8 = 8\/17<\/p>\n<p>Since, cosec&nbsp;\u03b8&nbsp;= 1\/sin \u03b8<\/p>\n<p>Also, cosec \u03b8 = Hypotenuse\/Perpendicular<\/p>\n<p>\u2234 cosec \u03b8 = 17\/8<\/p>\n<p>Since,&nbsp; cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos \u03b8 = 15\/17<\/p>\n<p>Since,&nbsp; sec \u03b8 = 1\/cos \u03b8<\/p>\n<p>Also,&nbsp; sec&nbsp;\u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234 sec \u03b8 = 17\/15<\/p>\n<p>Since, cot \u03b8 = 1\/tan \u03b8<\/p>\n<p>Also, &nbsp;cot \u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234 cot \u03b8 = 15\/8<\/p>\n<p><strong>(ix) cot \u03b8 = 12\/5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, cot&nbsp;\u03b8 = 12\/5 \u2026\u2026\u2026\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>cot \u03b8 = 1\/tan \u03b8<\/p>\n<p>cot \u03b8 = Base\/Perpendicular&nbsp;\u2026\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we&nbsp;have<\/p>\n<p>Base = 12 and Perpendicular side = 5<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of base (AB) and perpendicular (BC) to get the hypotenuse (AC),<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 12<sup>2<\/sup>&nbsp;+ 5<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>= 144 + 25<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 169<\/p>\n<p>AC =&nbsp;\u221a169<\/p>\n<p>AC = 13<\/p>\n<p>Hence, Hypotenuse = 13<\/p>\n<p>By definition,<\/p>\n<p>Since,&nbsp; sin \u03b8 = perpendicular\/Hypotenuse<\/p>\n<p>\u2234&nbsp;sin \u03b8= 5\/13<\/p>\n<p>Since, cosec&nbsp;\u03b8&nbsp;= 1\/sin \u03b8<\/p>\n<p>Also, cosec&nbsp;\u03b8= Hypotenuse\/Perpendicualar<\/p>\n<p>\u2234 cosec&nbsp;\u03b8&nbsp;= 13\/5<\/p>\n<p>Since,&nbsp; cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos \u03b8 = 12\/13<\/p>\n<p>Since,&nbsp; sec \u03b8 = 1\/cos\u03b8<\/p>\n<p>Also,&nbsp; sec&nbsp;\u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234 sec \u03b8 = 13\/12<\/p>\n<p>Since,&nbsp; tan\u03b8 = 1\/cot \u03b8<\/p>\n<p>Also,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234 tan \u03b8 = 5\/12<\/p>\n<p><strong>(x) &nbsp;sec \u03b8 = 13\/5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, sec \u03b8 = 13\/5\u2026\u2026.\u2026 (1)<\/p>\n<p>By definition,<\/p>\n<p>sec&nbsp;\u03b8 = Hypotenuse\/Base\u2026\u2026\u2026\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we&nbsp;get<\/p>\n<p>Base = 5 and&nbsp;Hypotenuse = 13<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>And putting the value of base side (AB) and hypotenuse (AC) to get the perpendicular side (BC),<\/p>\n<p>13<sup>2<\/sup>&nbsp;= 5<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 13<sup>2&nbsp;<\/sup>\u2013 5<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>=169 \u2013 25<\/p>\n<p>BC<sup>2<\/sup>= 144<\/p>\n<p>BC=&nbsp;\u221a144<\/p>\n<p>BC = 12<\/p>\n<p>Hence, Perpendicular = 12<\/p>\n<p>By definition,<\/p>\n<p>Since, &nbsp;sin \u03b8 = perpendicular\/Hypotenuse<\/p>\n<p>\u2234 sin \u03b8= 12\/13<\/p>\n<p>Since, cosec&nbsp;\u03b8= 1\/ sin \u03b8<\/p>\n<p>Also, cosec \u03b8= Hypotenuse\/Perpendicular<\/p>\n<p>\u2234&nbsp;cosec&nbsp;\u03b8&nbsp;= 13\/12<\/p>\n<p>Since,&nbsp; cos \u03b8= 1\/sec \u03b8<\/p>\n<p>Also,&nbsp; cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234 cos \u03b8 = 5\/13<\/p>\n<p>Since,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234&nbsp;tan \u03b8 = 12\/5<\/p>\n<p>Since,&nbsp; cot \u03b8 = 1\/tan \u03b8<\/p>\n<p>Also,&nbsp; cot \u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234 cot \u03b8 = 5\/12<\/p>\n<p><strong>(xi) &nbsp;cosec \u03b8 = \u221a10<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, cosec&nbsp;\u03b8&nbsp;= \u221a10\/1&nbsp; &nbsp;\u2026\u2026..\u2026 (1)<\/p>\n<p>By definition,<\/p>\n<p>cosec&nbsp;\u03b8 = Hypotenuse\/ Perpendicualar \u2026\u2026.\u2026.(2)<\/p>\n<p>And, cosec\u03b8 = 1\/sin \u03b8<\/p>\n<p>On comparing eq.(1) and(2), we get<\/p>\n<p>Perpendicular side = 1 and&nbsp;Hypotenuse =&nbsp;\u221a10<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base side (AB),<\/p>\n<p>(\u221a10)<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ 1<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>= (\u221a10)<sup>2<\/sup>&nbsp;\u2013 1<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>= 10 \u2013 1<\/p>\n<p>AB =&nbsp;\u221a9<\/p>\n<p>AB = 3<\/p>\n<p>So, Base side = 3<\/p>\n<p>By definition,<\/p>\n<p>Since,&nbsp; sin \u03b8 = Perpendicular\/Hypotenuse<\/p>\n<p>\u2234&nbsp;sin \u03b8 = 1\/\u221a10<\/p>\n<p>Since,&nbsp; cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos \u03b8 = 3\/\u221a10<\/p>\n<p>Since,&nbsp; sec \u03b8 = 1\/cos \u03b8<\/p>\n<p>Also, sec&nbsp;\u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234&nbsp;sec \u03b8 = \u221a10\/3<\/p>\n<p>Since,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234&nbsp;tan \u03b8 = 1\/3<\/p>\n<p>Since,&nbsp; cot \u03b8 = 1\/tan \u03b8<\/p>\n<p>\u2234 cot \u03b8 = 3\/1<\/p>\n<p><strong>(xii)&nbsp; cos \u03b8 =12\/15<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have;&nbsp;cos&nbsp;\u03b8 = 12\/15&nbsp;\u2026\u2026\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>cos&nbsp;\u03b8 = Base\/Hypotenuse\u2026\u2026\u2026 (2)<\/p>\n<p>By comparing eq. (1) and (2), we&nbsp;get;<\/p>\n<p>Base =12 and Hypotenuse = 15<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC, we get<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of base (AB) and hypotenuse (AC) to get the perpendicular (BC),<\/p>\n<p>15<sup>2<\/sup>&nbsp;= 12<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 15<sup>2<\/sup>&nbsp;\u2013 12<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 225 \u2013 144<\/p>\n<p>BC&nbsp;<sup>2<\/sup>= 81<\/p>\n<p>BC =&nbsp;\u221a81<\/p>\n<p>BC = 9<\/p>\n<p>So, Perpendicular = 9<\/p>\n<p>By definition,<\/p>\n<p>Since,&nbsp; sin \u03b8 = perpendicular\/Hypotenuse<\/p>\n<p>\u2234&nbsp;sin&nbsp;\u03b8 = 9\/15 = 3\/5<\/p>\n<p>Since, cosec&nbsp;\u03b8&nbsp;= 1\/sin&nbsp;\u03b8<\/p>\n<p>Also, cosec \u03b8 = Hypotenuse\/Perpendicular<\/p>\n<p>\u2234&nbsp;cosec&nbsp;\u03b8= 15\/9 = 5\/3<\/p>\n<p>Since,&nbsp; sec \u03b8 = 1\/cos \u03b8<\/p>\n<p>Also,&nbsp; sec&nbsp;\u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234 sec&nbsp;\u03b8 = 15\/12 = 5\/4<\/p>\n<p>Since,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234&nbsp;tan&nbsp;\u03b8 = 9\/12 = 3\/4<\/p>\n<p>Since,&nbsp; cot&nbsp;\u03b8 = 1\/tan&nbsp;\u03b8<\/p>\n<p>Also,&nbsp; cot \u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234 cot&nbsp;\u03b8 = 12\/9 = 4\/3<\/p>\n<p><strong>2. In a \u25b3 ABC, right-angled at B, AB = 24 cm , BC = 7 cm. Determine<\/strong><\/p>\n<p><strong>(i) sin A , cos A (ii) sin C, cos C<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-1.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 2\" width=\"308\" height=\"144\"><\/strong><\/p>\n<p><strong>(i)&nbsp;<\/strong>Given: In \u25b3ABC, AB = 24 cm, BC = 7cm and \u2220ABC&nbsp;= 90<sup>o<\/sup><\/p>\n<p>To find: sin A, cos A<\/p>\n<p>By using Pythagoras\u2019 theorem in \u25b3ABC, we have<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 24<sup>2<\/sup>&nbsp;+ 7<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 576 + 49<\/p>\n<p>AC<sup>2<\/sup>= 625<\/p>\n<p>AC =&nbsp;\u221a625<\/p>\n<p>AC= 25<\/p>\n<p>Hence, Hypotenuse = 25<\/p>\n<p>By definition,<\/p>\n<p>sin A = Perpendicular side opposite to angle A\/ Hypotenuse<\/p>\n<p>sin A = BC\/ AC<\/p>\n<p>sin A = 7\/ 25<\/p>\n<p>And,<\/p>\n<p>cos A = Base side adjacent to angle A\/Hypotenuse<\/p>\n<p>cos A = AB\/ AC<\/p>\n<p>cos A = 24\/ 25<\/p>\n<p><strong>&nbsp;<\/strong><\/p>\n<p><strong>(ii)&nbsp;<\/strong>Given:&nbsp;In \u25b3ABC , AB = 24 cm and BC = 7cm and \u2220ABC&nbsp;= 90<sup>o<\/sup><\/p>\n<p>To find: sin C, cos C<\/p>\n<p>By using Pythagoras\u2019 theorem in \u25b3ABC, we have<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 24<sup>2<\/sup>&nbsp;+ 7<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 576 + 49<\/p>\n<p>AC<sup>2<\/sup>= 625<\/p>\n<p>AC =&nbsp;\u221a625<\/p>\n<p>AC= 25<\/p>\n<p>Hence, Hypotenuse = 25<\/p>\n<p>By definition,<\/p>\n<p>sin C = Perpendicular side opposite to angle C\/Hypotenuse<\/p>\n<p>sin C = AB\/ AC<\/p>\n<p>sin C = 24\/ 25<\/p>\n<p>And,<\/p>\n<p>cos C = Base side adjacent to angle C\/Hypotenuse<\/p>\n<p>cos A = BC\/AC<\/p>\n<p>cos A = 7\/25<\/p>\n<p><strong>3. In fig. 5.37, find tan P and cot R. Is tan P = cot R?&nbsp;<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-2.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 3\" width=\"214\" height=\"190\"><\/strong><\/p>\n<p>By using Pythagoras theorem in \u25b3PQR, we have<\/p>\n<p>PR<sup>2<\/sup>&nbsp;= PQ<sup>2<\/sup>&nbsp;+ QR<sup>2<\/sup><\/p>\n<p>Putting the length of given side PR and PQ in the above equation,<\/p>\n<p>13<sup>2&nbsp;<\/sup>= 12<sup>2<\/sup>&nbsp;+ QR<sup>2<\/sup><\/p>\n<p>QR<sup>2<\/sup>&nbsp;= 13<sup>2<\/sup>&nbsp;\u2013 12<sup>2<\/sup><\/p>\n<p>QR<sup>2<\/sup>&nbsp;= 169 \u2013 144<\/p>\n<p>QR<sup>2&nbsp;<\/sup>= 25<\/p>\n<p>QR =&nbsp;\u221a25 = 5<\/p>\n<p>By definition,<\/p>\n<p>tan P = Perpendicular side opposite to P\/ Base side adjacent to angle P<\/p>\n<p>tan P = QR\/PQ<\/p>\n<p>tan P = 5\/12&nbsp;\u2026\u2026\u2026. (1)<\/p>\n<p>And,<\/p>\n<p>cot R= Base\/Perpendicular<\/p>\n<p>cot R= QR\/PQ<\/p>\n<p>cot R= 5\/12&nbsp;\u2026. (2)<\/p>\n<p>When comparing equation (1) and (2), we can see that R.H.S of both the equation is equal.<\/p>\n<p>Therefore, L.H.S of both equations should also be equal.<\/p>\n<p>\u2234 tan P = cot R<\/p>\n<p><strong>Yes, tan P = cot R =&nbsp;5\/12<\/strong><\/p>\n<p><strong>4. If sin A =&nbsp;9\/41, compute cos A and tan A.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-3.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 4\" width=\"191\" height=\"188\"><\/strong><\/p>\n<p>Given that,&nbsp;sin A = 9\/41&nbsp;\u2026\u2026\u2026\u2026. (1)<\/p>\n<p>Required to find: cos A, tan A<\/p>\n<p>By definition, we know that<\/p>\n<p>sin A = Perpendicular\/ Hypotenuse\u2026\u2026\u2026\u2026\u2026(2)<\/p>\n<p>On Comparing eq. (1) and (2), we get<\/p>\n<p>Perpendicular side = 9 and Hypotenuse = 41<\/p>\n<p>Let\u2019s construct \u25b3ABC&nbsp;as shown below,<\/p>\n<p>And, here the length of base AB is unknown.<\/p>\n<p>Thus, by using Pythagoras&#8217; theorem in \u25b3ABC, we get<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>41<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ 9<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 41<sup>2<\/sup>&nbsp;\u2013 9<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 168 \u2013 81<\/p>\n<p>AB= 1600<\/p>\n<p>AB =&nbsp;\u221a1600<\/p>\n<p>AB = 40<\/p>\n<p>\u21d2 Base of triangle ABC, AB = 40<\/p>\n<p>We know that,<\/p>\n<p>cos A = Base\/ Hypotenuse<\/p>\n<p>cos A =AB\/AC<\/p>\n<p>cos A =40\/41<\/p>\n<p>And,<\/p>\n<p>tan A = Perpendicular\/ Base<\/p>\n<p>tan A = BC\/AB<\/p>\n<p>tan A = 9\/40<\/p>\n<p><strong>5.&nbsp; Given 15cot A= 8, find sin A and sec A.<\/strong><\/p>\n<p><strong>Solution<\/strong><\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 5\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-4.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 5\" width=\"385\" height=\"180\"><\/p>\n<p>We have, 15cot A = 8<\/p>\n<p>Required to find: sin A and sec A<\/p>\n<p>As, 15 cot A = 8<\/p>\n<p>\u21d2 cot A = 8\/15 \u2026\u2026.(1)<\/p>\n<p>And we know,<\/p>\n<p>cot A = 1\/tan A<\/p>\n<p>Also by definition,<\/p>\n<p>Cot A = Base side adjacent to \u2220A\/ Perpendicular side opposite to \u2220A&nbsp;\u2026. (2)<\/p>\n<p>On comparing equations (1) and (2), we get<\/p>\n<p>The base side adjacent to \u2220A = 8<\/p>\n<p>The perpendicular side opposite to \u2220A = 15<\/p>\n<p>So, by using Pythagoras&#8217; theorem to \u25b3ABC, we have<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+BC<sup>2<\/sup><\/p>\n<p>Substituting values for sides from the figure<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 8<sup>2<\/sup>&nbsp;+ 15<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 64 + 225<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 289<\/p>\n<p>AC =&nbsp;\u221a289<\/p>\n<p>AC = 17<\/p>\n<p>Therefore, hypotenuse =17<\/p>\n<p>By definition,<\/p>\n<p>sin A = Perpendicular\/Hypotenuse<\/p>\n<p>\u21d2 sin A= BC\/AC<\/p>\n<p>sin A= 15\/17 (using values from the above)<\/p>\n<p>Also,<\/p>\n<p>sec A= 1\/ cos A<\/p>\n<p>\u21d2 secA = Hypotenuse\/ Base side adjacent to \u2220A<\/p>\n<p>\u2234 sec A= 17\/8<\/p>\n<p><strong>&nbsp;6. In \u25b3PQR, right-angled at Q, PQ = 4cm and RQ = 3 cm. Find the value of sin P, sin R, sec P, and sec R.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-5.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 6\" width=\"404\" height=\"189\"><\/strong><\/p>\n<p>Given:<\/p>\n<p>\u25b3PQR&nbsp;is right-angled at Q.<\/p>\n<p>PQ = 4cm<\/p>\n<p>RQ = 3cm<\/p>\n<p>Required to find: sin P, sin R, sec P, sec R<\/p>\n<p>Given \u25b3PQR,<\/p>\n<p>By using Pythagoras theorem to \u25b3PQR, we get<\/p>\n<p>PR<sup>2<\/sup>&nbsp;= PQ<sup>2<\/sup>&nbsp;+RQ<sup>2<\/sup><\/p>\n<p>Substituting the respective values,<\/p>\n<p>PR<sup>2<\/sup>&nbsp;= 4<sup>2<\/sup>&nbsp;+3<sup>2<\/sup><\/p>\n<p>PR<sup>2<\/sup>&nbsp;= 16 + 9<\/p>\n<p>PR<sup>2<\/sup>&nbsp;= 25<\/p>\n<p>PR =&nbsp;\u221a25<\/p>\n<p>PR = 5<\/p>\n<p>\u21d2 Hypotenuse =5<\/p>\n<p>By definition,<\/p>\n<p>sin P = Perpendicular side opposite to angle P\/ Hypotenuse<\/p>\n<p>sin P = RQ\/ PR<\/p>\n<p>\u21d2 sin P = 3\/5<\/p>\n<p>And,<\/p>\n<p>sin R = Perpendicular side opposite to angle R\/ Hypotenuse<\/p>\n<p>sin R = PQ\/ PR<\/p>\n<p>\u21d2 sin R = 4\/5<\/p>\n<p>And,<\/p>\n<p>sec P=1\/cos P<\/p>\n<p>secP = Hypotenuse\/ Base side adjacent to \u2220P<\/p>\n<p>sec P = PR\/ PQ<\/p>\n<p>\u21d2 sec P = 5\/4<\/p>\n<p>Now,<\/p>\n<p>sec R = 1\/cos R<\/p>\n<p>secR = Hypotenuse\/ Base side adjacent to \u2220R<\/p>\n<p>sec R = PR\/ RQ<\/p>\n<p>\u21d2 sec R = 5\/3<\/p>\n<p><strong>7.<\/strong>&nbsp;<strong>If&nbsp;cot \u03b8 = 7\/8, evaluate<\/strong><\/p>\n<p><strong>&nbsp;&nbsp;&nbsp;&nbsp;(i)&nbsp;&nbsp;&nbsp;(1+sin \u03b8)(1\u2013sin \u03b8)\/ (1+cos \u03b8)(1\u2013cos \u03b8)<\/strong><\/p>\n<p><strong>&nbsp;&nbsp;&nbsp;&nbsp;(ii) &nbsp;cot<sup>2&nbsp;<\/sup>\u03b8<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)&nbsp;<\/strong>Required to evaluate:<br><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 7\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5.gif\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 7\">, given = cot \u03b8 = 7\/8<\/p>\n<p>Taking the numerator, we have<\/p>\n<p>(1+sin \u03b8)(1\u2013sin \u03b8) = 1 \u2013 sin<sup>2<\/sup>&nbsp;\u03b8 [Since, (a+b)(a-b) = a<sup>2<\/sup>&nbsp;\u2013 b<sup>2<\/sup>]<\/p>\n<p>Similarly,<\/p>\n<p>(1+cos \u03b8)(1\u2013cos \u03b8) = 1 \u2013 cos<sup>2<\/sup>&nbsp;\u03b8<\/p>\n<p>We know that,<\/p>\n<p>sin<sup>2<\/sup>&nbsp;\u03b8 + cos<sup>2<\/sup>&nbsp;\u03b8 = 1<\/p>\n<p>\u21d2 1 \u2013 cos<sup>2<\/sup>&nbsp;\u03b8 = sin<sup>2<\/sup>&nbsp;\u03b8<\/p>\n<p>And,<\/p>\n<p>1 \u2013 sin<sup>2<\/sup>&nbsp;\u03b8 = cos<sup>2<\/sup>&nbsp;\u03b8<\/p>\n<p>Thus,<\/p>\n<p>(1+sin \u03b8)(1 \u2013sin \u03b8) = 1 \u2013 sin<sup>2<\/sup>&nbsp;\u03b8 = cos<sup>2<\/sup>&nbsp;\u03b8<\/p>\n<p>(1+cos \u03b8)(1\u2013cos \u03b8) = 1 \u2013 cos<sup>2<\/sup>&nbsp;\u03b8 = sin<sup>2<\/sup>&nbsp;\u03b8<\/p>\n<p>\u21d2<br><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 8\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-1.gif\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 8\"><\/p>\n<p>= cos<sup>2<\/sup>&nbsp;\u03b8\/ sin<sup>2<\/sup>&nbsp;\u03b8<\/p>\n<p>= (cos \u03b8\/sin \u03b8)<sup>2<\/sup><\/p>\n<p>And, we know that (cos \u03b8\/sin \u03b8) = cot \u03b8<\/p>\n<p>\u21d2<br><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 9\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-2.gif\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 9\"><\/p>\n<p><strong>=&nbsp;<\/strong>(cot \u03b8)<sup>2<\/sup><\/p>\n<p>= (7\/8)<sup>2<\/sup><\/p>\n<p>= 49\/ 64<\/p>\n<p><strong>(ii)&nbsp;<\/strong>Given,<\/p>\n<p>cot \u03b8 = 7\/8<\/p>\n<p>So, by squaring on both sides we get<\/p>\n<p>(cot \u03b8)<sup>2<\/sup>&nbsp;= (7\/8)<sup>2<\/sup><\/p>\n<p>\u2234 cot \u03b8<sup>2<\/sup>&nbsp;= 49\/64<\/p>\n<p><strong>8. If&nbsp;3cot A = 4, check whether&nbsp;(1\u2013tan<sup>2<\/sup>A)\/(1+tan<sup>2<\/sup>A) = (cos<sup>2<\/sup>A \u2013 sin<sup>2<\/sup>A)&nbsp;or not.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 10\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-6.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 10\" width=\"203\" height=\"183\"><\/strong><\/p>\n<p>Given,<\/p>\n<p>3cot A = 4<\/p>\n<p>\u21d2 cot A = 4\/3<\/p>\n<p>By definition,<\/p>\n<p>tan A = 1\/ Cot A = 1\/ (4\/3)<\/p>\n<p>\u21d2 tan A = 3\/4<\/p>\n<p>Thus,<\/p>\n<p>The base side adjacent to \u2220A = 4<\/p>\n<p>The perpendicular side opposite to \u2220A = 3<\/p>\n<p>In \u0394ABC, Hypotenuse is unknown.<\/p>\n<p>Thus, by applying Pythagoras&#8217; theorem in \u0394ABC,<\/p>\n<p>We get<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 4<sup>2<\/sup>&nbsp;+ 3<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 16 + 9<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 25<\/p>\n<p>AC = \u221a25<\/p>\n<p>AC = 5<\/p>\n<p>Hence, hypotenuse = 5<\/p>\n<p>Now, we can find that<\/p>\n<p>sin A = opposite side to \u2220A\/ Hypotenuse = 3\/5<\/p>\n<p>And,<\/p>\n<p>cos A = adjacent side to \u2220A\/ Hypotenuse = 4\/5<\/p>\n<p>Taking the LHS,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 11\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-7.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 11\"><\/p>\n<p>Thus, LHS = 7\/25<\/p>\n<p>Now, taking RHS,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 12\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-8.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 12\"><\/p>\n<p><strong>9. If tan \u03b8 = a\/b, find the value of (cos \u03b8 + sin \u03b8)\/ (cos \u03b8 \u2013 sin \u03b8)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>tan \u03b8 = a\/b<\/p>\n<p>And we know by definition that<\/p>\n<p>tan \u03b8 = opposite side\/ adjacent side<\/p>\n<p>Thus, by comparison,<\/p>\n<p>Opposite side = a and adjacent side = b<\/p>\n<p>To find the hypotenuse, we know that by Pythagoras&#8217; theorem that<\/p>\n<p>Hypotenuse<sup>2<\/sup>&nbsp;= opposite side<sup>2<\/sup>&nbsp;+ adjacent side<sup>2<\/sup><\/p>\n<p>\u21d2 Hypotenuse = \u221a(a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>)<\/p>\n<p>So, by definition<\/p>\n<p>sin \u03b8 = opposite side\/ Hypotenuse<\/p>\n<p>sin \u03b8 = a\/ \u221a(a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>)<\/p>\n<p>And,<\/p>\n<p>cos \u03b8 = adjacent side\/ Hypotenuse<\/p>\n<p>cos \u03b8 = b\/ \u221a(a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>)<\/p>\n<p>Now,<\/p>\n<p>After substituting for cos \u03b8 and sin \u03b8, we have<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-9.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 13\"><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 14\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-10.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 14\"><\/p>\n<p>\u2234<\/p>\n<p>Hence, proved.<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 15\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-11.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 15\"><\/p>\n<p><strong>10. If&nbsp;3 tan \u03b8 = 4, find the value of<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, 3 tan \u03b8 = 4<\/p>\n<p>\u21d2 tan \u03b8 = 4\/3<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 16\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-12.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 16\"><\/p>\n<p>From, let\u2019s divide the numerator and denominator by cos \u03b8.<\/p>\n<p>We get,<\/p>\n<p>(4 \u2013 tan \u03b8) \/ (2 + tan \u03b8)<\/p>\n<p>\u21d2 (4 \u2013 (4\/3)) \/ (2 + (4\/3)) [using the value of tan \u03b8]<\/p>\n<p>\u21d2 (12 \u2013 4) \/ (6 + 4) [After taking LCM and cancelling it]<\/p>\n<p>\u21d2 8\/10 = 4\/5<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-13.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 17\"><\/p>\n<p>\u2234 = 4\/5<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-14.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 18\"><\/p>\n<p><strong>11. If&nbsp;3 cot \u03b8 = 2, find the value of<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, 3 cot \u03b8 = 2<\/p>\n<p>\u21d2 cot \u03b8 = 2\/3<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 19\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-15.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 19\"><\/p>\n<p>From, let\u2019s divide the numerator and denominator by sin \u03b8.<\/p>\n<p>We get,<\/p>\n<p>(4 \u20133 cot \u03b8) \/ (2 + 6 cot \u03b8)<\/p>\n<p>\u21d2 (4 \u2013 3(2\/3)) \/ (2 + 6(2\/3)) [using the value of tan \u03b8]<\/p>\n<p>\u21d2 (4 \u2013 2) \/ (2 + 4) [After taking LCM and simplifying it]<\/p>\n<p>\u21d2 2\/6 = 1\/3<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 20\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-16.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 20\"><\/p>\n<p>\u2234 = 1\/3<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 21\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-17.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 21\"><\/p>\n<p><strong>12. If&nbsp;tan \u03b8 = a\/b, prove that<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, tan \u03b8 = a\/b<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 22\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-18.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 22\"><\/p>\n<p>From LHS, let\u2019s divide the numerator and denominator by cos \u03b8.<\/p>\n<p>And we get,<\/p>\n<p>(a tan \u03b8 \u2013 b) \/ (a tan \u03b8 + b)<\/p>\n<p>\u21d2 (a(a\/b) \u2013 b) \/ (a(a\/b) + b) [using the value of tan \u03b8]<\/p>\n<p>\u21d2 (a<sup>2<\/sup>&nbsp;\u2013 b<sup>2<\/sup>)\/b<sup>2<\/sup>&nbsp;\/ (a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>)\/b<sup>2<\/sup>&nbsp;[After taking LCM and simplifying it]<\/p>\n<p>\u21d2 (a<sup>2<\/sup>&nbsp;\u2013 b<sup>2<\/sup>)\/ (a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>)<\/p>\n<p>= RHS<\/p>\n<p>Hence, proved<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 23\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-19.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 23\"><\/p>\n<p><strong>13. If&nbsp;sec \u03b8 = 13\/5, show that<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 24\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-20.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 24\" width=\"212\" height=\"195\"><\/strong><\/p>\n<p>Given,<\/p>\n<p>sec \u03b8 = 13\/5<\/p>\n<p>We know that,<\/p>\n<p>sec \u03b8 = 1\/ cos \u03b8<\/p>\n<p>\u21d2 cos \u03b8 = 1\/ sec \u03b8 = 1\/ (13\/5)<\/p>\n<p>\u2234 cos \u03b8 = 5\/13 \u2026\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>cos \u03b8 = adjacent side\/ hypotenuse \u2026.. (2)<\/p>\n<p>Comparing (1) and (2), we have<\/p>\n<p>Adjacent side = 5 and hypotenuse = 13<\/p>\n<p>By Pythagoras theorem,<\/p>\n<p>Opposite side = \u221a((hypotenuse)<sup>&nbsp;2<\/sup>&nbsp;\u2013 (adjacent side)<sup>2<\/sup>)<\/p>\n<p>= \u221a(13<sup>2<\/sup>&nbsp;\u2013 5<sup>2<\/sup>)<\/p>\n<p>= \u221a(169 \u2013 25)<\/p>\n<p>= \u221a(144)<\/p>\n<p>= 12<\/p>\n<p>Thus, the opposite side = 12<\/p>\n<p>By definition,<\/p>\n<p>tan \u03b8 = opposite side\/ adjacent side<\/p>\n<p>\u2234 tan \u03b8 = 12\/ 5<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 25\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-21.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 25\"><\/p>\n<p>From, let\u2019s divide the numerator and denominator by cos \u03b8.<\/p>\n<p>We get,<\/p>\n<p>(2 tan \u03b8 \u2013 3) \/ (4 tan \u03b8 \u2013 9)<\/p>\n<p>\u21d2 (2(12\/5) \u2013 3) \/ (4(12\/5) \u2013 9) [using the value of tan \u03b8]<\/p>\n<p>\u21d2 (24 \u2013 15) \/ (48 \u2013 45) [After taking LCM and canceling it]<\/p>\n<p>\u21d2 9\/3 = 3<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 26\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-22.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 26\"><\/p>\n<p>\u2234 = 3<\/p>\n<p><strong>14. If&nbsp;cos \u03b8 = 12\/13, show that&nbsp;sin \u03b8(1 \u2013 tan \u03b8) = 35\/156<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 27\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-23.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 27\" width=\"254\" height=\"212\"><\/p>\n<p>Given, cos \u03b8 = 12\/13\u2026\u2026 (1)<\/p>\n<p>By definition, we know that<\/p>\n<p>cos \u03b8 = Base side adjacent to \u2220\u03b8 \/ Hypotenuse\u2026\u2026. (2)<\/p>\n<p>When comparing equation (1) and (2), we get<\/p>\n<p>Base side adjacent to&nbsp;\u2220\u03b8&nbsp;= 12 and Hypotenuse = 13<\/p>\n<p>From the figure,<\/p>\n<p>Base side BC = 12<\/p>\n<p>Hypotenuse AC = 13<\/p>\n<p>Side AB is unknown here, and it can be found by using Pythagoras&#8217; theorem.<\/p>\n<p>Thus, by applying Pythagoras&#8217; theorem,<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>13<sup>2&nbsp;<\/sup>= AB<sup>2<\/sup>&nbsp;+ 12<sup>2<\/sup><\/p>\n<p>Therefore,<\/p>\n<p>AB<sup>2&nbsp;<\/sup>= 13<sup>2&nbsp;<\/sup>\u2013 12<sup>2<\/sup><\/p>\n<p>AB<sup>2&nbsp;<\/sup>= 169 \u2013 144<\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 25<\/p>\n<p>AB =&nbsp;\u221a25<\/p>\n<p>AB = 5 \u2026. (3)<\/p>\n<p>Now, we know that<\/p>\n<p>sin \u03b8 = Perpendicular side opposite to \u2220\u03b8 \/ Hypotenuse<\/p>\n<p>Thus, sin \u03b8 = AB\/AC [from figure]<\/p>\n<p>\u21d2 sin \u03b8 = 5\/13\u2026 (4)<\/p>\n<p>And, tan \u03b8 = sin \u03b8 \/ cos \u03b8 = (5\/13) \/ (12\/13)<\/p>\n<p>\u21d2 tan \u03b8 = 12\/13\u2026 (5)<\/p>\n<p>Taking L.H.S we have<\/p>\n<p>L.H.S =&nbsp;sin \u03b8 (1 \u2013 tan \u03b8)<\/p>\n<p>Substituting the value of sin \u03b8 and tan \u03b8 from equations (4) and (5),<\/p>\n<p>We get,<\/p>\n<p><strong><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 29\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-25.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 29\"><\/strong><\/p>\n<p><strong>15.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 30\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-26.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 30\" width=\"204\" height=\"180\"><\/p>\n<p>Given, cot \u03b8 = 1\/<strong>\u221a<\/strong>3\u2026\u2026. (1)<\/p>\n<p>By definition, we know that,<\/p>\n<p>cot \u03b8 = 1\/ tan \u03b8<\/p>\n<p>And, since tan \u03b8 = perpendicular side opposite to \u2220\u03b8&nbsp;\/ Base side adjacent to \u2220\u03b8<\/p>\n<p>\u21d2 cot \u03b8 = Base side adjacent to \u2220\u03b8&nbsp;\/ perpendicular side opposite to \u2220\u03b8&nbsp;\u2026\u2026 (2)<\/p>\n<p>[Since they are reciprocal to each other]<\/p>\n<p>On comparing equations (1) and (2), we get<\/p>\n<p>The base side adjacent to \u2220\u03b8 = 1 and the Perpendicular side opposite to \u2220\u03b8 = \u221a3<\/p>\n<p>Therefore, the triangle formed is<\/p>\n<p>On substituting the values of known sides as AB = \u221a3 and BC = 1,<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= (\u221a3) + 1<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 3 + 1<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= 4<\/p>\n<p>AC = \u221a4<\/p>\n<p>Therefore, AC = 2 \u2026 &nbsp;&nbsp;(3)<\/p>\n<p>Now, by definition,<\/p>\n<p>sin \u03b8 = Perpendicular side opposite to \u2220\u03b8 \/ Hypotenuse = AB \/ AC<\/p>\n<p>\u21d2 sin \u03b8 = \u221a3\/ 2 \u2026\u2026(4)<\/p>\n<p>And, cos \u03b8 = Base side adjacent to \u2220\u03b8 \/ Hypotenuse = BC \/ AC<\/p>\n<p>\u21d2 cos \u03b8 = 1\/ 2 \u2026.. (5)<\/p>\n<p>Now, taking L.H.S, we have<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 31\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-27.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 31\"><\/p>\n<p>Substituting the values from equation (4) and (5), we have<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 33\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-29.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 33\"><\/p>\n<p><strong>16.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, tan \u03b8 = 1\/ \u221a7 \u2026..(1)<\/p>\n<p><strong><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 34\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-30.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 34\" width=\"417\" height=\"195\"><\/strong><\/p>\n<p>By definition, we know that<\/p>\n<p>tan \u03b8 = Perpendicular side opposite to \u2220\u03b8 \/ Base side adjacent to \u2220\u03b8 \u2026\u2026(2)<\/p>\n<p>On comparing equations (1) and (2), we have<\/p>\n<p>The perpendicular side opposite to \u2220\u03b8 = 1<\/p>\n<p>The base side adjacent to \u2220\u03b8 = \u221a7<\/p>\n<p>Thus, the triangle representing&nbsp;\u2220 \u03b8&nbsp;is,<\/p>\n<p>Hypotenuse AC is unknown, and it can be found by using Pythagoras&#8217; theorem.<\/p>\n<p>By applying Pythagoras&#8217; theorem, we have<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>AC<sup>2&nbsp;<\/sup>= 1<sup>2<\/sup>&nbsp;+ (\u221a7)<sup>2<\/sup><\/p>\n<p>AC&nbsp;<sup>2<\/sup>&nbsp;= 1 + 7<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 8<\/p>\n<p>AC = \u221a8<\/p>\n<p>\u21d2 AC = 2\u221a2<\/p>\n<p>By definition,<\/p>\n<p>sin \u03b8 = Perpendicular side opposite to \u2220\u03b8 \/ Hypotenuse = AB \/ AC<\/p>\n<p>\u21d2 sin \u03b8 = 1\/ 2\u221a2<\/p>\n<p>And, since cosec \u03b8 = 1\/sin \u03b8<\/p>\n<p>\u21d2 cosec \u03b8 = 2\u221a2 \u2026\u2026.. (3)<\/p>\n<p>Now,<\/p>\n<p>cos \u03b8 = Base side adjacent to \u2220\u03b8 \/ Hypotenuse = BC \/ AC<\/p>\n<p>\u21d2 cos \u03b8 = \u221a7\/ 2\u221a2<\/p>\n<p>And, since sec \u03b8 = 1\/ sin \u03b8<\/p>\n<p>\u21d2 sec \u03b8 = 2\u221a2\/ \u221a7 \u2026\u2026. (4)<\/p>\n<p>Taking the L.H.S of the equation,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 35\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-31.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 35\"><\/p>\n<p>Substituting the value of cosec \u03b8 and sec \u03b8 from equations (3) and (4), we get<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 37\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-33.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 37\"><\/p>\n<p><strong>17. If&nbsp;sec \u03b8 = 5\/4, find the value of<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>sec \u03b8 = 5\/4<\/p>\n<p>We know that,<\/p>\n<p>sec \u03b8 = 1\/ cos \u03b8<\/p>\n<p>\u21d2 cos \u03b8 = 1\/ (5\/4) = 4\/5 \u2026\u2026 (1)<\/p>\n<p>By definition,<\/p>\n<p>cos \u03b8 = Base side adjacent to \u2220\u03b8 \/ Hypotenuse \u2026. (2)<\/p>\n<p>On comparing equations (1) and (2), we have<\/p>\n<p>Hypotenuse&nbsp;= 5<\/p>\n<p>The base side adjacent to \u2220\u03b8 = 4<\/p>\n<p>Thus, the triangle representing&nbsp;\u2220 \u03b8&nbsp;is ABC.<\/p>\n<p>The perpendicular side opposite to \u2220\u03b8, AB is unknown, and it can be found by using Pythagoras&#8217; theorem.<\/p>\n<p>By applying Pythagoras\u2019 theorem, we have<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>AB<sup>2&nbsp;<\/sup>= AC<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 5<sup>2<\/sup>&nbsp;\u2013 4<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 25 \u2013 16<\/p>\n<p>AB = \u221a9<\/p>\n<p>\u21d2 AB = 3<\/p>\n<p>By definition,<\/p>\n<p>sin \u03b8 = Perpendicular side opposite to \u2220\u03b8 \/ Hypotenuse = AB \/ AC<\/p>\n<p>\u21d2 sin \u03b8 = 3\/ 5 \u2026..(3)<\/p>\n<p>Now, tan \u03b8 = Perpendicular side opposite to \u2220\u03b8 \/ Base side adjacent to \u2220\u03b8<\/p>\n<p>\u21d2 tan \u03b8 = 3\/ 4 \u2026\u2026(4)<\/p>\n<p>And, since cot \u03b8 = 1\/ tan \u03b8<\/p>\n<p>\u21d2 cot \u03b8 = 4\/ 3 \u2026\u2026(5)<\/p>\n<p>Now,<\/p>\n<p>Substituting the value of sin \u03b8, cos \u03b8, cot \u03b8, and tan \u03b8 from the equations (1), (3), (4), and (5), we have<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 38\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-34.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 38\"><\/p>\n<p>= 12\/7<\/p>\n<p>Therefore,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 39\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-35.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 39\"><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 40\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-36.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 40\"><\/p>\n<p><strong>18. If&nbsp;tan \u03b8 = 12\/13, find the value of<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>tan \u03b8 = 12\/13 \u2026\u2026.. (1)<\/p>\n<p>We know that by definition,<\/p>\n<p>tan \u03b8 = Perpendicular side opposite to \u2220\u03b8 \/ Base side adjacent to \u2220\u03b8 \u2026\u2026 (2)<\/p>\n<p>On comparing equations (1) and (2), we have<\/p>\n<p>The perpendicular side opposite to \u2220\u03b8 = 12<\/p>\n<p>The base side adjacent to \u2220\u03b8 = 13<\/p>\n<p>Thus, in the triangle representing \u2220 \u03b8, we have,<\/p>\n<p>Hypotenuse AC is the unknown, and it can be found by using Pythagoras&#8217; theorem.<\/p>\n<p>So, by applying Pythagoras\u2019 theorem, we have<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= 12<sup>2<\/sup>&nbsp;+ 13<sup>2<\/sup><\/p>\n<p>AC&nbsp;<sup>2<\/sup>&nbsp;= 144 + 169<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 313<\/p>\n<p>\u21d2 AC = \u221a313<\/p>\n<p>By definition,<\/p>\n<p>sin \u03b8 = Perpendicular side opposite to \u2220\u03b8 \/ Hypotenuse = AB \/ AC<\/p>\n<p>\u21d2 sin \u03b8 = 12\/ \u221a313\u2026..(3)<\/p>\n<p>And, cos \u03b8 = Base side adjacent to \u2220\u03b8 \/ Hypotenuse = BC \/ AC<\/p>\n<p>\u21d2 cos \u03b8 = 13\/ \u221a313 \u2026..(4)<\/p>\n<p>Now, substituting the value of sin \u03b8 and cos \u03b8 from equations (3) and (4), respectively, in the equation below,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 41\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-37.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 41\"><\/p>\n<p>Therefore,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 42\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-38.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 42\"><\/p>\n<hr>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-for-class-10-maths-chapter-5-exercise-52-page-no-541\"><\/span>RD Sharma Solutions for Class 10 Maths Chapter 5 Exercise 5.2 Page No: 5.41<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Evaluate each of the following:<\/strong><\/p>\n<p><strong>1. sin&nbsp;45<sup>\u2218<\/sup>&nbsp;sin&nbsp;30<sup>\u2218<\/sup>&nbsp;+ cos&nbsp;45<sup>\u2218<\/sup>&nbsp;cos&nbsp;30<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-39.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 1\"><\/p>\n<p><strong>2. sin&nbsp;60<sup>\u2218<\/sup>&nbsp;cos&nbsp;30<sup>\u2218<\/sup>&nbsp;+ cos&nbsp;60<sup>\u2218<\/sup>&nbsp;sin&nbsp;30<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-40.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 2\"><\/p>\n<p><strong>3. &nbsp;cos&nbsp;60<sup>\u2218<\/sup>&nbsp;cos&nbsp;45<sup>\u2218<\/sup>&nbsp;\u2013 sin&nbsp;60<sup>\u2218<\/sup>&nbsp;sin&nbsp;45<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-41.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 3\"><\/p>\n<p><strong>4. sin<sup>2&nbsp;<\/sup>30<sup>\u2218<\/sup>&nbsp;+ sin<sup>2&nbsp;<\/sup>45<sup>\u2218<\/sup>&nbsp;+ sin<sup>2&nbsp;<\/sup>60<sup>\u2218<\/sup>&nbsp;+ sin<sup>2&nbsp;<\/sup>90<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-42.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 4\"><\/p>\n<p><strong>5. &nbsp;cos<sup>2&nbsp;<\/sup>30<sup>\u2218<\/sup>&nbsp;+ cos<sup>2<\/sup>&nbsp;45<sup>\u2218<\/sup>&nbsp;+ cos<sup>2&nbsp;<\/sup>60<sup>\u2218<\/sup>&nbsp;+ cos<sup>2&nbsp;<\/sup>90<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 5\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-43.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 5\"><\/p>\n<p><strong>6. tan<sup>2&nbsp;<\/sup>30<sup>\u2218<\/sup>&nbsp;+ tan<sup>2<\/sup>&nbsp;45<sup>\u2218<\/sup>&nbsp;+ tan<sup>2<\/sup>&nbsp;60<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-44.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 6\"><\/p>\n<p><strong>7. 2sin<sup>2<\/sup>&nbsp;30<sup>\u2218<\/sup>&nbsp;\u2212 3cos<sup>2&nbsp;<\/sup>45<sup>\u2218<\/sup>&nbsp;+ tan<sup>2&nbsp;<\/sup>60<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 7\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-45.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 7\"><\/p>\n<p><strong>8. sin<sup>2<\/sup>&nbsp;30<sup>\u2218<\/sup>&nbsp;cos<sup>2<\/sup>45<sup>\u2218<\/sup>&nbsp;+ 4tan<sup>2<\/sup>&nbsp;30<sup>\u2218<\/sup>&nbsp;+ (1\/2) sin<sup>2<\/sup>&nbsp;90<sup>\u2218<\/sup>&nbsp;\u2212 2cos<sup>2<\/sup>&nbsp;90<sup>\u2218<\/sup>&nbsp;+ (1\/24) cos20<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 8\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-46.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 8\"><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 9\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-47.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 9\"><\/p>\n<p><strong>9. 4(sin<sup>4<\/sup>&nbsp;60<sup>\u2218<\/sup>&nbsp;+ cos<sup>4&nbsp;<\/sup>30<sup>\u2218<\/sup>) \u2212 3(tan<sup>2&nbsp;<\/sup>60<sup>\u2218&nbsp;<\/sup>\u2212 tan<sup>2<\/sup>&nbsp;45<sup>\u2218<\/sup>) + 5cos<sup>2<\/sup>&nbsp;45<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 10\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-48.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 10\"><\/p>\n<p><strong>10. (cosec<sup>2<\/sup>&nbsp;45<sup>\u2218<\/sup>&nbsp;sec<sup>2&nbsp;<\/sup>30<sup>\u2218<\/sup>)(sin<sup>2<\/sup>&nbsp;30<sup>\u2218<\/sup>&nbsp;+ 4cot<sup>2<\/sup>&nbsp;45<sup>\u2218<\/sup>&nbsp;\u2212 sec<sup>2<\/sup>&nbsp;60<sup>\u2218<\/sup>)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 11\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/12\/r-d-sharma-solutions-for-class-10-maths-chapter-5-49.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 11\"><\/p>\n<p><strong>11. &nbsp;cosec<sup>3<\/sup>&nbsp;30<sup>\u2218<\/sup>&nbsp;cos60<sup>\u2218<\/sup>&nbsp;tan<sup>3<\/sup>&nbsp;45<sup>\u2218<\/sup>&nbsp;sin<sup>2<\/sup>&nbsp;90<sup>\u2218<\/sup>&nbsp;sec<sup>2<\/sup>&nbsp;45<sup>\u2218<\/sup>&nbsp;cot30<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 12\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-50.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 12\"><\/p>\n<p><strong>12. cot<sup>2<\/sup>&nbsp;30<sup>\u2218<\/sup>&nbsp;\u2212 2cos<sup>2<\/sup>&nbsp;60<sup>\u2218<\/sup>&nbsp;\u2212 (3\/4)sec<sup>2<\/sup>&nbsp;45<sup>\u2218<\/sup>&nbsp;\u2013 4sec<sup>2<\/sup>&nbsp;30<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Using trigonometric values, we have<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/09\/r-d-sharma-solutions-for-class-10-maths-chapter-5-51.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 13\"><\/p>\n<p><strong>13. (cos0<sup>\u2218<\/sup>&nbsp;+ sin45<sup>\u2218<\/sup>&nbsp;+ sin30<sup>\u2218<\/sup>)(sin90<sup>\u2218<\/sup>&nbsp;\u2212 cos45<sup>\u2218<\/sup>&nbsp;+ cos60<sup>\u2218<\/sup>)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(cos0<sup>\u2218<\/sup>&nbsp;+ sin45<sup>\u2218<\/sup>&nbsp;+ sin30<sup>\u2218<\/sup>)(sin90<sup>\u2218<\/sup>&nbsp;\u2212 cos45<sup>\u2218<\/sup>&nbsp;+ cos60<sup>\u2218<\/sup>)<\/p>\n<p>Using trigonometric values, we have<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 14\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-52.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 14\"><\/p>\n<p><strong><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 15\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-53.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 15\"><\/strong><\/p>\n<p><strong>15. 4\/cot<sup>2<\/sup>&nbsp;30<sup>\u2218<\/sup>&nbsp;+ 1\/sin<sup>2<\/sup>&nbsp;60<sup>\u2218<\/sup>&nbsp;\u2212 cos<sup>2<\/sup>&nbsp;45<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 16\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-54.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 16\"><\/p>\n<p><strong>16. 4(sin<sup>4&nbsp;<\/sup>30<sup>\u2218<\/sup>&nbsp;+ cos<sup>2<\/sup>&nbsp;60<sup>\u2218<\/sup>) \u2212 3(cos<sup>2<\/sup>&nbsp;45<sup>\u2218<\/sup>&nbsp;\u2212 sin<sup>2<\/sup>&nbsp;90<sup>\u2218<\/sup>) \u2212 sin<sup>2<\/sup>&nbsp;60<sup>\u2218<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Using trigonometric values, we have<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-55.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 17\"><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-56.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 18\"><\/p>\n<p><strong>17.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Using trigonometric values, we have<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 19\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-57.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 19\"><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 20\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-58.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 20\"><strong>18.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Using trigonometric values, we have<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 21\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-59.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 21\"><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 22\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-60.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 22\"><\/p>\n<p><strong>19.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Using trigonometric values, we have<\/p>\n<p><strong><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 23\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-61.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.2 - 23\"><\/strong><\/p>\n<p><strong>Find the value of x in each of the following: (20-25)<\/strong><\/p>\n<p><strong>20. 2sin 3x = \u221a3<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>2 sin 3x = \u221a3<\/p>\n<p>sin 3x = \u221a3\/2<\/p>\n<p>sin 3x = sin 60\u00b0<\/p>\n<p>3x = 60\u00b0<\/p>\n<p>x = 20\u00b0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-for-class-10-maths-chapter-5-exercise-53-page-no-552\"><\/span>RD Sharma Solutions for Class 10 Maths Chapter 5 Exercise 5.3 Page No: 5.52<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Evaluate the following:<\/strong><\/p>\n<p><strong>(i) sin 20<sup>o<\/sup>\/ cos 70<sup>o<\/sup><\/strong><\/p>\n<p><strong>(ii) cos 19<sup>o<\/sup>\/ sin 71<sup>o<\/sup><\/strong><\/p>\n<p><strong>(iii) sin 21<sup>o<\/sup>\/ cos 69<sup>o<\/sup><\/strong><\/p>\n<p><strong>(iv) tan 10<sup>o<\/sup>\/ cot 80<sup>o<\/sup><\/strong><\/p>\n<p><strong>(v) sec 11<sup>o<\/sup>\/ cosec 79<sup>o<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) We have,<\/p>\n<p>sin 20<sup>o<\/sup>\/ cos 70<sup>o<\/sup>&nbsp;= sin (90<sup>o<\/sup>&nbsp;\u2013 70<sup>o<\/sup>)\/ cos 70<sup>o&nbsp;<\/sup>= cos 70<sup>o<\/sup>\/ cos70<sup>o<\/sup>&nbsp;= 1 [\u2235 sin (90 \u2013 \u03b8) = cos \u03b8]<\/p>\n<p>(ii) We have,<\/p>\n<p>cos 19<sup>o<\/sup>\/ sin 71<sup>o&nbsp;<\/sup>= cos (90<sup>o<\/sup>&nbsp;\u2013 71<sup>o<\/sup>)\/ sin 71<sup>o&nbsp;<\/sup>= sin 71<sup>o<\/sup>\/ sin 71<sup>o<\/sup>&nbsp;= 1 [\u2235 cos (90 \u2013 \u03b8) = sin \u03b8]<\/p>\n<p>(iii) We have,<\/p>\n<p>sin 21<sup>o<\/sup>\/ cos 69<sup>o&nbsp;<\/sup>= sin (90<sup>o<\/sup>&nbsp;\u2013 69<sup>o<\/sup>)\/ cos 69<sup>o<\/sup>&nbsp;= cos 69<sup>o<\/sup>\/ cos69<sup>o<\/sup>&nbsp;= 1 [\u2235 sin (90 \u2013 \u03b8) = cos \u03b8]<\/p>\n<p>(iv) We have,<\/p>\n<p>tan 10<sup>o<\/sup>\/ cot 80<sup>o&nbsp;<\/sup>= tan (90<sup>o<\/sup>&nbsp;\u2013 10<sup>o<\/sup>) \/ cot 80<sup>o<\/sup>&nbsp;= cot 80<sup>o<\/sup>\/ cos80<sup>o<\/sup>&nbsp;= 1 [\u2235 tan (90 \u2013 \u03b8) = cot \u03b8]<\/p>\n<p>(v) We have,<\/p>\n<p>sec 11<sup>o<\/sup>\/ cosec 79<sup>o<\/sup>&nbsp;= sec (90<sup>o<\/sup>&nbsp;\u2013 79<sup>o<\/sup>)\/ cosec 79<sup>o<\/sup>&nbsp;= cosec 79<sup>o<\/sup>\/ cosec 79<sup>o<\/sup>&nbsp;= 1<\/p>\n<p>[\u2235 sec (90 \u2013 \u03b8) = cosec \u03b8]<\/p>\n<p><strong>2. Evaluate the following:<\/strong><\/p>\n<p><strong><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-62.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 1\"><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, [\u2235 sin (90 \u2013 \u03b8) = cos \u03b8 and cos (90 \u2013 \u03b8) = sin \u03b8]<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-63.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 2\"><\/p>\n<p>= 1<sup>2<\/sup>&nbsp;+ 1<sup>2<\/sup>&nbsp;= 1 + 1<\/p>\n<p>= 2<\/p>\n<p><strong>(ii) cos 48\u00b0- sin 42\u00b0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that, cos (90\u00b0 \u2212 \u03b8) = sin \u03b8.<\/p>\n<p>So,<\/p>\n<p>cos 48\u00b0&nbsp;\u2013&nbsp;sin 42\u00b0&nbsp;=&nbsp;cos (90\u00b0 \u2212 42\u00b0) \u2013 sin 42\u00b0&nbsp;= sin 42\u00b0&nbsp;\u2013&nbsp;sin 42\u00b0= 0<\/p>\n<p>Thus, the value of cos 48\u00b0 \u2013 sin 42\u00b0 is 0.<\/p>\n<p>&nbsp;<\/p>\n<p><strong><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-64.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 3\"><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, [\u2235 cot (90 \u2013 \u03b8) = tan \u03b8 and cos (90 \u2013 \u03b8) = sin \u03b8]<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-65.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 4\"><\/p>\n<p>= 1 \u2013 1\/2(1)<\/p>\n<p>= 1\/2<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 5\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-66.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 5\"><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, [\u2235 sin (90 \u2013 \u03b8) = cos \u03b8 and cos (90 \u2013 \u03b8) = sin \u03b8]<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-67.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 6\"><\/p>\n<p>= 1 \u2013 1<\/p>\n<p>= 0<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-68.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 6\"><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have [\u2235 cot (90 \u2013 \u03b8) = tan \u03b8 and tan (90 \u2013 \u03b8) = cot \u03b8]<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 8\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-69.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 8\"><\/p>\n<p>= tan (90<sup>o<\/sup>&nbsp;\u2013 35<sup>o<\/sup>)\/ cot 55<sup>o<\/sup>&nbsp;+ cot (90<sup>o<\/sup>&nbsp;\u2013 12<sup>o<\/sup>)\/ tan 12<sup>o<\/sup>&nbsp;\u2013 1<\/p>\n<p>= cot 55<sup>o<\/sup>\/ cot 55<sup>o<\/sup>&nbsp;+ tan 12<sup>o<\/sup>\/ tan 12<sup>o<\/sup>&nbsp;\u2013 1<\/p>\n<p>= 1 + 1 \u2013 1<\/p>\n<p>= 1<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 9\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-70.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 9\"><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p>We have [\u2235 sin (90 \u2013 \u03b8) = cos \u03b8 and sec (90 \u2013 \u03b8) = cosec \u03b8]<\/p>\n<p><strong><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 10\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-71.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 10\"><\/strong><\/p>\n<p>= sec (90<sup>o<\/sup>&nbsp;\u2013 20<sup>o<\/sup>)\/ cosec 20<sup>o<\/sup>&nbsp;+ sin (90<sup>o<\/sup>&nbsp;\u2013 31<sup>o<\/sup>)\/ cos 31<sup>o<\/sup><\/p>\n<p>= cosec 20<sup>o<\/sup>\/ cosec 20<sup>o<\/sup>&nbsp;+ cos 12<sup>o<\/sup>\/ cos 12<sup>o<\/sup><\/p>\n<p>= 1 + 1<\/p>\n<p>= 2<\/p>\n<p><strong>(vii)&nbsp;cosec 31\u00b0 \u2013 sec 59\u00b0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have,<\/p>\n<p>cosec 31\u00b0 \u2013 sec 59\u00b0<\/p>\n<p>Since, cosec (90 \u2013 \u03b8) = cos \u03b8<\/p>\n<p>So,<\/p>\n<p>cosec 31\u00b0 \u2013 sec 59\u00b0 = cosec (90\u00b0 \u2013 59<sup>o<\/sup>) \u2013 sec 59\u00b0 = sec 59\u00b0 \u2013 sec 59\u00b0 = 0<\/p>\n<p>Thus,<\/p>\n<p>cosec 31\u00b0 \u2013 sec 59\u00b0 = 0<\/p>\n<p><strong>(viii) (sin 72\u00b0 + cos 18\u00b0) (sin 72\u00b0 \u2013 cos 18\u00b0)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that,<\/p>\n<p>sin (90 \u2013 \u03b8) = cos \u03b8<\/p>\n<p>So, the given can be expressed as<\/p>\n<p>(sin 72\u00b0 + cos 18\u00b0) (sin (90 \u2013 18)\u00b0 \u2013 cos 18\u00b0)<\/p>\n<p>= (sin 72\u00b0 + cos 18\u00b0) (cos 18\u00b0 \u2013 cos 18\u00b0)<\/p>\n<p>= (sin 72\u00b0 + cos 18\u00b0) x 0<\/p>\n<p>= 0<\/p>\n<p><strong>(ix) sin 35\u00b0 sin 55\u00b0 \u2013 cos 35\u00b0 cos 55\u00b0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that,<\/p>\n<p>sin (90 \u2013 \u03b8) = cos \u03b8<\/p>\n<p>So, the given can be expressed as<\/p>\n<p>sin (90 \u2013 55)\u00b0 sin (90 \u2013 35)\u00b0 \u2013 cos 35\u00b0 cos 55\u00b0<\/p>\n<p>= cos 55\u00b0 cos 35\u00b0 \u2013 cos 35\u00b0 cos 55\u00b0<\/p>\n<p>= 0<\/p>\n<p><strong>(x) tan 48\u00b0 tan 23\u00b0 tan 42\u00b0 tan 67\u00b0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that,<\/p>\n<p>tan (90 \u2013 \u03b8) = cot \u03b8<\/p>\n<p>So, the given can be expressed as<\/p>\n<p>tan (90 \u2013 42)\u00b0 tan (90 \u2013 67)\u00b0 tan 42\u00b0 tan 67\u00b0<\/p>\n<p>= cot 42\u00b0 cot 67\u00b0 tan 42\u00b0 tan 67\u00b0<\/p>\n<p>= (cot 42\u00b0 tan 42\u00b0)(cot 67\u00b0 tan 67\u00b0)<\/p>\n<p>= 1 x 1 [\u2235 tan \u03b8 x cot \u03b8 = 1]<\/p>\n<p>= 1<\/p>\n<p><strong>(xi) sec 50\u00b0 sin 40\u00b0 + cos 40\u00b0 cosec 50\u00b0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that,<\/p>\n<p>sin (90 \u2013 \u03b8) = cos \u03b8 and cos (90 \u2013 \u03b8) = sin \u03b8<\/p>\n<p>So, the given can be expressed as<\/p>\n<p>sec 50\u00b0 sin (90 \u2013 50)\u00b0 + cos (90 \u2013 50)\u00b0 cosec 50\u00b0<\/p>\n<p>= sec 50\u00b0 cos 50\u00b0 + sin 50\u00b0 cosec 50\u00b0<\/p>\n<p>= 1 + 1 [\u2235 sin \u03b8 x cosec \u03b8 = 1 and cos \u03b8 x sec \u03b8 = 1]<\/p>\n<p>= 2<\/p>\n<p><strong>3. Express each one of the following in terms of trigonometric ratios of angles lying between 0<sup>o<\/sup>&nbsp;and 45<sup>o<\/sup><\/strong><\/p>\n<p><strong>(i) sin 59<sup>o<\/sup>&nbsp;+ cos 56<sup>o<\/sup>&nbsp;(ii) tan 65<sup>o<\/sup>&nbsp;+ cot 49<sup>o<\/sup>&nbsp;(iii) sec 76<sup>o<\/sup>&nbsp;+ cosec 52<sup>o<\/sup><\/strong><\/p>\n<p><strong>(iv) cos 78<sup>o<\/sup>&nbsp;+ sec 78<sup>o<\/sup>&nbsp;(v) cosec 54<sup>o<\/sup>&nbsp;+ sin 72<sup>o<\/sup>&nbsp;(vi) cot 85<sup>o<\/sup>&nbsp;+ cos 75<sup>o<\/sup><\/strong><\/p>\n<p><strong>(vii) sin 67<sup>o<\/sup>&nbsp;+ cos 75<sup>o<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Using the below trigonometric ratios of complementary angles, we find the required.<\/p>\n<p>sin (90 \u2013 \u03b8) = cos \u03b8 cosec (90 \u2013 \u03b8) = sec \u03b8<\/p>\n<p>cos (90 \u2013 \u03b8) = sin \u03b8 sec (90 \u2013 \u03b8) = cosec \u03b8<\/p>\n<p>tan (90 \u2013 \u03b8) = cot \u03b8 cot (90 \u2013 \u03b8) = tan \u03b8<\/p>\n<p>(i) sin 59<sup>o<\/sup>&nbsp;+ cos 56<sup>o<\/sup>&nbsp;= sin (90 \u2013 31)<sup>o<\/sup>&nbsp;+ cos (90 \u2013 34)<sup>o<\/sup>&nbsp;= cos 31<sup>o<\/sup>&nbsp;+ sin 34<sup>o<\/sup><\/p>\n<p>(ii) tan 65<sup>o<\/sup>&nbsp;+ cot 49<sup>o<\/sup>&nbsp;= tan (90 \u2013 25)<sup>o<\/sup>&nbsp;+ cot (90 -41)<sup>o<\/sup>&nbsp;= cot 25<sup>o<\/sup>&nbsp;+ tan 41<sup>o<\/sup><\/p>\n<p>(iii) sec 76<sup>o<\/sup>&nbsp;+ cosec 52<sup>o<\/sup>&nbsp;= sec (90 \u2013 14)<sup>o<\/sup>&nbsp;+ cosec (90 \u2013 38)<sup>o<\/sup>&nbsp;= cosec 14<sup>o<\/sup>&nbsp;+ sec 38<sup>o<\/sup><\/p>\n<p>(iv) cos 78<sup>o<\/sup>&nbsp;+ sec 78<sup>o&nbsp;<\/sup>= cos (90 \u2013 12)<sup>o<\/sup>&nbsp;+ sec (90 \u2013 12)<sup>o<\/sup>&nbsp;= sin 12<sup>o<\/sup>&nbsp;+ cosec 12<sup>o<\/sup><\/p>\n<p>(v) cosec 54<sup>o<\/sup>&nbsp;+ sin 72<sup>o<\/sup>&nbsp;= cosec (90 \u2013 36)<sup>o<\/sup>&nbsp;+ sin (90 \u2013 18)<sup>o<\/sup>&nbsp;= sec 36<sup>o<\/sup>&nbsp;+ cos 18<sup>o<\/sup><\/p>\n<p>(vi) cot 85<sup>o<\/sup>&nbsp;+ cos 75<sup>o<\/sup>&nbsp;= cot (90 \u2013 5)<sup>o<\/sup>&nbsp;+ cos (90 \u2013 15)<sup>o<\/sup>&nbsp;= tan 5<sup>o<\/sup>&nbsp;+ sin 15<sup>o<\/sup><\/p>\n<p><strong>4. Express cos 75<sup>o<\/sup>&nbsp;+ cot 75<sup>o<\/sup>&nbsp;in terms of angles between 0<sup>o<\/sup>&nbsp;and 30<sup>o<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>cos 75<sup>o<\/sup>&nbsp;+ cot 75<sup>o<\/sup><\/p>\n<p>Since, cos (90 \u2013 \u03b8) = sin \u03b8 and cot (90 \u2013 \u03b8) = tan \u03b8<\/p>\n<p>cos 75<sup>o<\/sup>&nbsp;+ cot 75<sup>o&nbsp;<\/sup>= cos (90 \u2013 15)<sup>o<\/sup>&nbsp;+ cot (90 \u2013 15)<sup>o<\/sup>&nbsp;= sin 15<sup>o&nbsp;<\/sup>+ tan 15<sup>o<\/sup><\/p>\n<p>Hence, cos 75<sup>o<\/sup>&nbsp;+ cot 75<sup>o<\/sup>&nbsp;can be expressed as sin 15<sup>o&nbsp;<\/sup>+ tan 15<sup>o<\/sup><\/p>\n<p><strong>5. If sin 3A = cos (A \u2013 26<sup>o<\/sup>), where 3A is an acute angle, find the value of A.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>sin 3A = cos (A \u2013 26<sup>o<\/sup>)<\/p>\n<p>Using cos (90 \u2013 \u03b8) = sin \u03b8, we have<\/p>\n<p>sin 3A = sin (90<sup>o<\/sup>&nbsp;\u2013 (A \u2013 26<sup>o<\/sup>))<\/p>\n<p>Now, comparing both L.H.S. and R.H.S.,<\/p>\n<p>3A = 90<sup>o<\/sup>&nbsp;\u2013 (A \u2013 26<sup>o<\/sup>)<\/p>\n<p>3A + (A \u2013 26<sup>o<\/sup>) = 90<sup>o<\/sup><\/p>\n<p>4A \u2013 26<sup>o<\/sup>&nbsp;= 90<sup>o<\/sup><\/p>\n<p>4A = 116<sup>o<\/sup><\/p>\n<p>A = 116<sup>o<\/sup>\/4<\/p>\n<p>\u2234 A = 29<sup>o<\/sup><\/p>\n<p><strong>6. If A, B, and C are the interior angles of a triangle ABC, prove that<\/strong><\/p>\n<p><strong>(i) tan ((C + A)\/ 2) = cot (B\/2) (ii) sin ((B + C)\/ 2) = cos (A\/2)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that in triangle ABC the sum of the angles, i.e., A + B + C = 180<sup>o<\/sup><\/p>\n<p>So, C + A = 180<sup>o<\/sup>&nbsp;\u2013 B \u21d2 (C + A)\/2 = 90<sup>o<\/sup>&nbsp;\u2013 B\/2 \u2026\u2026 (i)<\/p>\n<p>And, B + C = 180<sup>o&nbsp;<\/sup>\u2013 A \u21d2 (B + C)\/2 = 90<sup>o<\/sup>&nbsp;\u2013 A\/2 \u2026\u2026. (ii)<\/p>\n<p>(i) L.H.S = tan ((C + A)\/ 2)<\/p>\n<p>\u21d2 tan ((C + A)\/ 2) = tan (90<sup>o<\/sup>&nbsp;\u2013 B\/2) [From (i)]<\/p>\n<p>= cot (B\/2) [\u2235 tan (90 \u2013 \u03b8) = cot \u03b8]<\/p>\n<p>= R.H.S<\/p>\n<ul>\n<li>Hence, proved<\/li>\n<\/ul>\n<p>(ii) L.H.S = sin ((B + C)\/2)<\/p>\n<p>\u21d2 sin ((B + C)\/ 2) = sin (90<sup>o<\/sup>&nbsp;\u2013 A\/2) [From (ii)]<\/p>\n<p>= cos (A\/2)<\/p>\n<p>= R.H.S.<\/p>\n<ul>\n<li>Hence, proved<\/li>\n<\/ul>\n<p><strong>7. Prove that:<\/strong><\/p>\n<p><strong>(i) tan 20\u00b0 tan 35\u00b0 tan 45\u00b0 tan 55\u00b0 tan 70\u00b0 = 1<\/strong><\/p>\n<p><strong>(ii) sin 48\u00b0 sec 48\u00b0 + cos 48\u00b0 cosec 42\u00b0 = 2<\/strong><\/p>\n<p><strong><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 11\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-72.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 11\"><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Taking L.H.S = tan 20\u00b0 tan 35\u00b0 tan 45\u00b0 tan 55\u00b0 tan 70\u00b0<\/p>\n<p>= tan (90\u00b0 \u2212 70\u00b0) tan (90\u00b0 \u2212 55\u00b0) tan 45\u00b0tan 55\u00b0 tan70\u00b0<\/p>\n<p>= cot 70\u00b0cot 55\u00b0 tan 45\u00b0 tan 55\u00b0 tan 70\u00b0&nbsp; [\u2235 tan (90 \u2013 \u03b8) = cot \u03b8]<\/p>\n<p>= (tan 70\u00b0cot 70\u00b0)(tan 55\u00b0cot 55\u00b0) tan 45\u00b0&nbsp; [\u2235 tan \u03b8 x cot \u03b8 = 1]<\/p>\n<p>= 1 \u00d7 1 \u00d7 1 = 1<\/p>\n<ul>\n<li>Hence, proved<\/li>\n<\/ul>\n<p>(ii) Taking L.H.S = sin 48\u00b0 sec 48\u00b0 + cos 48\u00b0 cosec 42\u00b0<\/p>\n<p>= sin 48\u00b0 sec (90\u00b0 \u2212 48\u00b0) + cos 48\u00b0 cosec (90\u00b0 \u2212 48\u00b0)<\/p>\n<p>[\u2235sec (90 \u2013 \u03b8) = cosec \u03b8 and cosec (90 \u2013 \u03b8) = sec \u03b8]<\/p>\n<p>= sin 48\u00b0cosec 48\u00b0 + cos 48\u00b0sec 48\u00b0&nbsp; [\u2235 cosec \u03b8 x sin \u03b8 = 1 and cos \u03b8 x sec \u03b8 = 1]<\/p>\n<p>= 1 + 1 = 2<\/p>\n<ul>\n<li>Hence, proved<\/li>\n<\/ul>\n<p>(iii) Taking the L.H.S,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 12\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/09\/r-d-sharma-solutions-for-class-10-maths-chapter-5-73.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 12\"><\/p>\n<p>= 1 + 1 \u2013 2<\/p>\n<p>= 2 \u2013 2<\/p>\n<p>= 0<\/p>\n<ul>\n<li>Hence, proved<\/li>\n<\/ul>\n<p>(iv) Taking L.H.S,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-74.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 13\"><\/p>\n<p>= 1 + 1<\/p>\n<p>= 2<\/p>\n<ul>\n<li>Hence, proved<\/li>\n<\/ul>\n<p><strong>8. Prove the following:<\/strong><\/p>\n<p><strong>(i) sin\u03b8 sin (90<sup>o<\/sup>&nbsp;\u2013 \u03b8) \u2013 cos \u03b8 cos (90<sup>o<\/sup>&nbsp;\u2013 \u03b8) = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Taking the L.H.S,<\/p>\n<p>sin\u03b8 sin (90<sup>o<\/sup>&nbsp;\u2013 \u03b8) \u2013 cos \u03b8 cos (90<sup>o<\/sup>&nbsp;\u2013 \u03b8)<\/p>\n<p>= sin \u03b8 cos \u03b8 \u2013 cos \u03b8 sin \u03b8 [\u2235 sin (90 \u2013 \u03b8) = cos \u03b8 and cos (90 \u2013 \u03b8) = sin \u03b8]<\/p>\n<p>= 0<\/p>\n<p><strong>(ii)<img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 14\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-75.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 14\"><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Taking the L.H.S,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 15\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-76.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 15\"><\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 16\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-77.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 16\">[\u2235 cosec \u03b8 x sin \u03b8 = 1 and cos \u03b8 x sec \u03b8 = 1]<\/p>\n<p>= 1 + 1<\/p>\n<p>= 2 = R.H.S<\/p>\n<ul>\n<li>Hence, proved<\/li>\n<\/ul>\n<p><strong>(iii)<img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 17\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-78.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 17\"><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Taking the L.H.S., [\u2235 tan (90<sup>o<\/sup>&nbsp;\u2013 \u03b8) = cot \u03b8]<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-79.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 18\"><\/p>\n<p>= 0 = R.H.S.<\/p>\n<ul>\n<li>Hence, proved<\/li>\n<\/ul>\n<p><strong>(iv)<img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 19\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-80.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 19\"><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Taking L.H.S., [\u2235 sin (90 \u2013 \u03b8) = cos \u03b8 and cos (90 \u2013 \u03b8) = sin \u03b8]<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 20\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5-81.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 20\"><\/p>\n<p>= sin<sup>2<\/sup>&nbsp;A = R.H.S.<\/p>\n<ul>\n<li>Hence, proved<\/li>\n<\/ul>\n<p><strong>(v) sin (50<sup>o<\/sup>&nbsp;+&nbsp;<\/strong>\u03b8<strong>) \u2013 cos (40<sup>o<\/sup>&nbsp;\u2013&nbsp;<\/strong>\u03b8<strong>) + tan 1<sup>o<\/sup>&nbsp;tan 10<sup>o<\/sup>&nbsp;tan 20<sup>o<\/sup>&nbsp;tan 70<sup>o<\/sup>&nbsp;tan 80<sup>o<\/sup>&nbsp;tan 89<sup>o<\/sup>&nbsp;= 1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Taking the L.H.S.,<\/p>\n<p>= sin (50<sup>o<\/sup>&nbsp;+ \u03b8) \u2013 cos (40<sup>o<\/sup>&nbsp;\u2013 \u03b8) + tan 1<sup>o<\/sup>&nbsp;tan 10<sup>o<\/sup>&nbsp;tan 20<sup>o<\/sup>&nbsp;tan 70<sup>o<\/sup>&nbsp;tan 80<sup>o<\/sup>&nbsp;tan 89<sup>o<\/sup><\/p>\n<p>= [sin (90<sup>o<\/sup>&nbsp;\u2013 (40<sup>o<\/sup>&nbsp;\u2013 \u03b8))]&nbsp;\u2013 cos (40<sup>o<\/sup>&nbsp;\u2013 \u03b8) + tan (90 \u2013 89)<sup>o<\/sup>&nbsp;tan (90 \u2013 80)<sup>o<\/sup>&nbsp;tan (90 \u2013 70)<sup>o<\/sup>&nbsp;tan 70<sup>o<\/sup>&nbsp;tan 80<sup>o<\/sup>&nbsp;tan 89<sup>o&nbsp;<\/sup>[\u2235 sin (90 \u2013 \u03b8) = cos \u03b8]<\/p>\n<p>= cos (40<sup>o<\/sup>&nbsp;\u2013 \u03b8) \u2013 cos (40<sup>o<\/sup>&nbsp;\u2013 \u03b8) + cot 89<sup>o<\/sup>&nbsp;cot 80<sup>o<\/sup>&nbsp;cot 70<sup>o<\/sup>&nbsp;tan 70<sup>o<\/sup>&nbsp;tan 80<sup>o<\/sup>&nbsp;tan 89<sup>o<\/sup><\/p>\n<p>[\u2235 tan (90<sup>o<\/sup>&nbsp;\u2013 \u03b8) = cot \u03b8]<\/p>\n<p>= 0 + (cot 89<sup>o<\/sup>&nbsp;x tan 89<sup>o<\/sup>) (cot 80<sup>o<\/sup>&nbsp;x tan 80<sup>o<\/sup>) (cot 70<sup>o<\/sup>&nbsp;x tan 70<sup>o<\/sup>)<\/p>\n<p>= 0 + 1 x 1 x 1 [\u2235 tan \u03b8 x cot \u03b8 = 1]<\/p>\n<p>= 1= R.H.S<\/p>\n<ul>\n<li>Hence, proved<\/li>\n<\/ul>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>&nbsp;Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-5-exercise-51\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631191349255\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-using-rd-sharma-class-10-maths-solutions-chapter-5-exercise-51\"><\/span>What are the benefits of using RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>1. Correct answers according to the latest CBSE guidelines and syllabus.<br \/>2. The RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.1 is written in simple language to assist students in their board examination, &amp; competitive examination preparation.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631191462667\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-class-10-maths-solutions-chapter-5-exercise-51-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.1 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.1 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631191556823\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-important-to-study-all-of-the-concepts-included-in-rd-sharma-class-10-solutions-chapter-5-exercise-51\"><\/span>Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, learning all of the concepts included in RD Sharma Solutions for Class 10 Maths Chapter 5 Exercise 5.1 is required in order to achieve high scores on the Class 10 board exams. These solutions were created by subject matter specialists who compiled model questions that covered all of the textbook\u2019s exercise questions.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1:&nbsp;This exercise includes problems that require you to determine all of the trigonometric ratios when only one is given. The RD Sharma Class 10 Solutions has all of the solutions to this and other chapters. Students can get RD Sharma Solutions for Class 10 Maths Chapter 5 &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-exercise-5-1\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 5 Exercise 5.1 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":126042,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126033"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126033"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126033\/revisions"}],"predecessor-version":[{"id":505671,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126033\/revisions\/505671"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126042"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126033"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126033"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126033"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}