{"id":126031,"date":"2023-09-12T18:22:00","date_gmt":"2023-09-12T12:52:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126031"},"modified":"2023-11-30T10:18:20","modified_gmt":"2023-11-30T04:48:20","slug":"rd-sharma-class-10-solutions-chapter-5-exercise-5-3","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-exercise-5-3\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126044\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-5-Exercise-5.3.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-5-Exercise-5.3.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-5-Exercise-5.3-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3:\u00a0<\/strong>The following RD Sharma Solutions Class 10 exercise 5.3 contains a number of problems to help students better understand this concept. The solutions are written in simple Language to meet the needs of all students. For a full explanation of the questions in this exercise, students can download <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Solutions for Class 10 Maths<\/strong><\/a> Chapter 5 Trigonometric Ratios Exercise 5.3 PDF, which is provided below.<\/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-69d04733e902b\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-exercise-5-3\/#where-can-i-download-rd-sharma-class-10-maths-solutions-chapter-5-exercise-53-free-pdf\" title=\"Where can I download RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.3 free PDF?\">Where can I download RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.3 free PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-exercise-5-3\/#is-it-important-to-study-all-of-the-concepts-included-in-rd-sharma-class-10-solutions-chapter-5-exercise-53\" title=\"Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3?\">Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-5-exercise-53-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/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-5.3.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-5.3.pdf\">RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-5-exercise-53-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3- Important Question with Answers<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>\u00a0= sin (90<sup>o<\/sup>\u00a0\u2013 70<sup>o<\/sup>)\/ cos 70<sup>o\u00a0<\/sup>= cos 70<sup>o<\/sup>\/ cos70<sup>o<\/sup>\u00a0= 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\u00a0<\/sup>= cos (90<sup>o<\/sup>\u00a0\u2013 71<sup>o<\/sup>)\/ sin 71<sup>o\u00a0<\/sup>= sin 71<sup>o<\/sup>\/ sin 71<sup>o<\/sup>\u00a0= 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\u00a0<\/sup>= sin (90<sup>o<\/sup>\u00a0\u2013 69<sup>o<\/sup>)\/ cos 69<sup>o<\/sup>\u00a0= cos 69<sup>o<\/sup>\/ cos69<sup>o<\/sup>\u00a0= 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\u00a0<\/sup>= tan (90<sup>o<\/sup>\u00a0\u2013 10<sup>o<\/sup>) \/ cot 80<sup>o<\/sup>\u00a0= cot 80<sup>o<\/sup>\/ cos80<sup>o<\/sup>\u00a0= 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>\u00a0= sec (90<sup>o<\/sup>\u00a0\u2013 79<sup>o<\/sup>)\/ cosec 79<sup>o<\/sup>\u00a0= cosec 79<sup>o<\/sup>\/ cosec 79<sup>o<\/sup>\u00a0= 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>\u00a0+ 1<sup>2<\/sup>\u00a0= 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\u00a0\u2013\u00a0sin 42\u00b0\u00a0=\u00a0cos (90\u00b0 \u2212 42\u00b0) \u2013 sin 42\u00b0\u00a0= sin 42\u00b0\u00a0\u2013\u00a0sin 42\u00b0= 0<\/p>\n<p>Thus, the value of cos 48\u00b0 \u2013 sin 42\u00b0 is 0.<\/p>\n<p>\u00a0<\/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>\u00a0\u2013 35<sup>o<\/sup>)\/ cot 55<sup>o<\/sup>\u00a0+ cot (90<sup>o<\/sup>\u00a0\u2013 12<sup>o<\/sup>)\/ tan 12<sup>o<\/sup>\u00a0\u2013 1<\/p>\n<p>= cot 55<sup>o<\/sup>\/ cot 55<sup>o<\/sup>\u00a0+ tan 12<sup>o<\/sup>\/ tan 12<sup>o<\/sup>\u00a0\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>\u00a0<\/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>\u00a0\u2013 20<sup>o<\/sup>)\/ cosec 20<sup>o<\/sup>\u00a0+ sin (90<sup>o<\/sup>\u00a0\u2013 31<sup>o<\/sup>)\/ cos 31<sup>o<\/sup><\/p>\n<p>= cosec 20<sup>o<\/sup>\/ cosec 20<sup>o<\/sup>\u00a0+ cos 12<sup>o<\/sup>\/ cos 12<sup>o<\/sup><\/p>\n<p>= 1 + 1<\/p>\n<p>= 2<\/p>\n<p><strong>(vii)\u00a0cosec 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>\u00a0and 45<sup>o<\/sup><\/strong><\/p>\n<p><strong>(i) sin 59<sup>o<\/sup>\u00a0+ cos 56<sup>o<\/sup>\u00a0(ii) tan 65<sup>o<\/sup>\u00a0+ cot 49<sup>o<\/sup>\u00a0(iii) sec 76<sup>o<\/sup>\u00a0+ cosec 52<sup>o<\/sup><\/strong><\/p>\n<p><strong>(iv) cos 78<sup>o<\/sup>\u00a0+ sec 78<sup>o<\/sup>\u00a0(v) cosec 54<sup>o<\/sup>\u00a0+ sin 72<sup>o<\/sup>\u00a0(vi) cot 85<sup>o<\/sup>\u00a0+ cos 75<sup>o<\/sup><\/strong><\/p>\n<p><strong>(vii) sin 67<sup>o<\/sup>\u00a0+ 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>\u00a0+ cos 56<sup>o<\/sup>\u00a0= sin (90 \u2013 31)<sup>o<\/sup>\u00a0+ cos (90 \u2013 34)<sup>o<\/sup>\u00a0= cos 31<sup>o<\/sup>\u00a0+ sin 34<sup>o<\/sup><\/p>\n<p>(ii) tan 65<sup>o<\/sup>\u00a0+ cot 49<sup>o<\/sup>\u00a0= tan (90 \u2013 25)<sup>o<\/sup>\u00a0+ cot (90 -41)<sup>o<\/sup>\u00a0= cot 25<sup>o<\/sup>\u00a0+ tan 41<sup>o<\/sup><\/p>\n<p>(iii) sec 76<sup>o<\/sup>\u00a0+ cosec 52<sup>o<\/sup>\u00a0= sec (90 \u2013 14)<sup>o<\/sup>\u00a0+ cosec (90 \u2013 38)<sup>o<\/sup>\u00a0= cosec 14<sup>o<\/sup>\u00a0+ sec 38<sup>o<\/sup><\/p>\n<p>(iv) cos 78<sup>o<\/sup>\u00a0+ sec 78<sup>o\u00a0<\/sup>= cos (90 \u2013 12)<sup>o<\/sup>\u00a0+ sec (90 \u2013 12)<sup>o<\/sup>\u00a0= sin 12<sup>o<\/sup>\u00a0+ cosec 12<sup>o<\/sup><\/p>\n<p>(v) cosec 54<sup>o<\/sup>\u00a0+ sin 72<sup>o<\/sup>\u00a0= cosec (90 \u2013 36)<sup>o<\/sup>\u00a0+ sin (90 \u2013 18)<sup>o<\/sup>\u00a0= sec 36<sup>o<\/sup>\u00a0+ cos 18<sup>o<\/sup><\/p>\n<p>(vi) cot 85<sup>o<\/sup>\u00a0+ cos 75<sup>o<\/sup>\u00a0= cot (90 \u2013 5)<sup>o<\/sup>\u00a0+ cos (90 \u2013 15)<sup>o<\/sup>\u00a0= tan 5<sup>o<\/sup>\u00a0+ sin 15<sup>o<\/sup><\/p>\n<p><strong>4. Express cos 75<sup>o<\/sup>\u00a0+ cot 75<sup>o<\/sup>\u00a0in terms of angles between 0<sup>o<\/sup>\u00a0and 30<sup>o<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>cos 75<sup>o<\/sup>\u00a0+ 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>\u00a0+ cot 75<sup>o\u00a0<\/sup>= cos (90 \u2013 15)<sup>o<\/sup>\u00a0+ cot (90 \u2013 15)<sup>o<\/sup>\u00a0= sin 15<sup>o\u00a0<\/sup>+ tan 15<sup>o<\/sup><\/p>\n<p>Hence, cos 75<sup>o<\/sup>\u00a0+ cot 75<sup>o<\/sup>\u00a0can be expressed as sin 15<sup>o\u00a0<\/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>\u00a0\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>\u00a0\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>\u00a0= 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, 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>\u00a0\u2013 B \u21d2 (C + A)\/2 = 90<sup>o<\/sup>\u00a0\u2013 B\/2 \u2026\u2026 (i)<\/p>\n<p>And, B + C = 180<sup>o\u00a0<\/sup>\u2013 A \u21d2 (B + C)\/2 = 90<sup>o<\/sup>\u00a0\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>\u00a0\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>\u00a0\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\u00a0 [\u2235 tan (90 \u2013 \u03b8) = cot \u03b8]<\/p>\n<p>= (tan 70\u00b0cot 70\u00b0)(tan 55\u00b0cot 55\u00b0) tan 45\u00b0\u00a0 [\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\u00a0 [\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>\u00a0\u2013 \u03b8) \u2013 cos \u03b8 cos (90<sup>o<\/sup>\u00a0\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>\u00a0\u2013 \u03b8) \u2013 cos \u03b8 cos (90<sup>o<\/sup>\u00a0\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><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\" \/>(ii)<\/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><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\" \/>(iii)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Taking the L.H.S, [\u2235 tan (90<sup>o<\/sup>\u00a0\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><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\" \/>(iv)<\/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>\u00a0A = R.H.S<\/p>\n<ul>\n<li>Hence Proved<\/li>\n<\/ul>\n<p><strong>(v) sin (50<sup>o<\/sup>\u00a0+\u00a0<\/strong>\u03b8<strong>) \u2013 cos (40<sup>o<\/sup>\u00a0\u2013\u00a0<\/strong>\u03b8<strong>) + tan 1<sup>o<\/sup>\u00a0tan 10<sup>o<\/sup>\u00a0tan 20<sup>o\u00a0<\/sup>tan 70<sup>o<\/sup>\u00a0tan 80<sup>o<\/sup>\u00a0tan 89<sup>o<\/sup>\u00a0= 1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Taking the L.H.S,<\/p>\n<p>= sin (50<sup>o<\/sup>\u00a0+ \u03b8) \u2013 cos (40<sup>o<\/sup>\u00a0\u2013 \u03b8) + tan 1<sup>o<\/sup>\u00a0tan 10<sup>o<\/sup>\u00a0tan 20<sup>o<\/sup>\u00a0tan 70<sup>o<\/sup>\u00a0tan 80<sup>o<\/sup>\u00a0tan 89<sup>o<\/sup><\/p>\n<p>= [sin (90<sup>o<\/sup>\u00a0\u2013 (40<sup>o<\/sup>\u00a0\u2013 \u03b8))]\u00a0\u2013 cos (40<sup>o<\/sup>\u00a0\u2013 \u03b8) + tan (90 \u2013 89)<sup>o<\/sup>\u00a0tan (90 \u2013 80)<sup>o<\/sup>\u00a0tan (90 \u2013 70)<sup>o<\/sup>\u00a0tan 70<sup>o<\/sup>\u00a0tan 80<sup>o<\/sup>\u00a0tan 89<sup>o\u00a0<\/sup>[\u2235 sin (90 \u2013 \u03b8) = cos \u03b8]<\/p>\n<p>= cos (40<sup>o<\/sup>\u00a0\u2013 \u03b8) \u2013 cos (40<sup>o<\/sup>\u00a0\u2013 \u03b8) + cot 89<sup>o<\/sup>\u00a0cot 80<sup>o<\/sup>\u00a0cot 70<sup>o<\/sup>\u00a0tan 70<sup>o<\/sup>\u00a0tan 80<sup>o<\/sup>\u00a0tan 89<sup>o<\/sup><\/p>\n<p>[\u2235 tan (90<sup>o<\/sup>\u00a0\u2013 \u03b8) = cot \u03b8]<\/p>\n<p>= 0 + (cot 89<sup>o<\/sup>\u00a0x tan 89<sup>o<\/sup>) (cot 80<sup>o<\/sup>\u00a0x tan 80<sup>o<\/sup>) (cot 70<sup>o<\/sup>\u00a0x 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 5.3. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 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-53\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3<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-1631191370300\" 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-53\"><\/span>What are the benefits of using RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.3?<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 5.3 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-1631191475149\" 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-53-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.3 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 5.3 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631191572143\" 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-53\"><\/span>Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3?<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 5.3 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 5.3:\u00a0The following RD Sharma Solutions Class 10 exercise 5.3 contains a number of problems to help students better understand this concept. The solutions are written in simple Language to meet the needs of all students. For a full explanation of the questions in this exercise, students can &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-exercise-5-3\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":126044,"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\/126031"}],"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=126031"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126031\/revisions"}],"predecessor-version":[{"id":514498,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126031\/revisions\/514498"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126044"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126031"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126031"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126031"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}