{"id":118576,"date":"2021-08-24T20:09:44","date_gmt":"2021-08-24T14:39:44","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=118576"},"modified":"2021-08-25T17:16:37","modified_gmt":"2021-08-25T11:46:37","slug":"rd-sharma-solutions-class-12-chapter-4-exercise-4-7","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-7\/","title":{"rendered":"RD Sharma Solutions For Class 12 Maths Exercise 4.7 Chapter 4 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-118753\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-12-Chapter-4-Exercise-4.7.png\" alt=\"RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-12-Chapter-4-Exercise-4.7.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-12-Chapter-4-Exercise-4.7-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7:\u00a0<\/strong>Students can utilise the PDF <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\"><strong>RD Sharma Solutions<\/strong><\/a> to quickly clear their doubts and solve questions according to the most recent exam pattern. It allows students to assess their understanding of the concepts covered in each of Chapter 4&#8217;s exercises. Students can download <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-4\/\"><strong>RD Sharma Solutions Class 12 Maths Chapter 4 Inverse Trigonometric Functions<\/strong><\/a> Exercise 4.7 from the links provided below to improve their performance on the board exam. The properties of inverse trigonometric functions are the focus of this exercise.<\/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-69e7cc4dab74b\" 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\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-7\/#how-many-questions-are-there-in-rd-sharma-solutions-class-12-chapter-4-exercise-47\" title=\"How many questions are there in RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7?\">How many questions are there in RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7?<\/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-solutions-class-12-chapter-4-exercise-4-7\/#is-rd-sharma-solutions-class-12-chapter-4-exercise-47-for-free\" title=\"Is RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7 for free?\">Is RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7 for free?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-7\/#where-can-i-download-rd-sharma-solutions-class-12-chapter-4-exercise-47-free-pdf\" title=\"Where can I download RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7 free PDF?\">Where can I download RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7 free PDF?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-12-chapter-4-exercise-47-free-pdf\"><\/span>Download RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7 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\/08\/RD-_Sharma-_Class_12_Chapter_04_4.7.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-_Sharma-_Class_12_Chapter_04_4.7.pdf\">RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-maths-rd-sharma-solutions-for-class-12-chapter-4-%e2%80%93-inverse-trigonometric-functions-exercise-47-important-questions-with-solution\"><\/span>Access answers to Maths RD Sharma Solutions For Class 12 Chapter 4 \u2013 Inverse Trigonometric Functions Exercise 4.7 Important Questions With Solution<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Evaluate each of the following:<\/strong><\/p>\n<p><strong>(i) sin<sup>-1<\/sup>(sin \u03c0\/6)<\/strong><\/p>\n<p><strong>(ii) sin<sup>-1<\/sup>(sin 7\u03c0\/6)<\/strong><\/p>\n<p><strong>(iii) sin<sup>-1<\/sup>(sin 5\u03c0\/6)<\/strong><\/p>\n<p><strong>(iv) sin<sup>-1<\/sup>(sin 13\u03c0\/7)<\/strong><\/p>\n<p><strong>(v) sin<sup>-1<\/sup>(sin 17\u03c0\/8)<\/strong><\/p>\n<p><strong>(vi) sin<sup>-1<\/sup>{(sin \u2013 17\u03c0\/8)}<\/strong><\/p>\n<p><strong>(vii) sin<sup>-1<\/sup>(sin 3)<\/strong><\/p>\n<p><strong>(viii) sin<sup>-1<\/sup>(sin 4)<\/strong><\/p>\n<p><strong>(ix) sin<sup>-1<\/sup>(sin 12)<\/strong><\/p>\n<p><strong>(x) sin<sup>-1<\/sup>(sin 2)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given sin<sup>-1<\/sup>(sin \u03c0\/6)<\/p>\n<p>We know that the value of sin \u03c0\/6 is \u00bd<\/p>\n<p>By substituting this value in sin<sup>-1<\/sup>(sin \u03c0\/6)<\/p>\n<p>We get, sin<sup>-1<\/sup>\u00a0(1\/2)<\/p>\n<p>Now let y = sin<sup>-1<\/sup>\u00a0(1\/2)<\/p>\n<p>Sin (\u03c0\/6) = \u00bd<\/p>\n<p>The range of principal value of sin<sup>-1<\/sup>(-\u03c0\/2, \u03c0\/2) and sin (\u03c0\/6) = \u00bd<\/p>\n<p>Therefore sin<sup>-1<\/sup>(sin \u03c0\/6) = \u03c0\/6<\/p>\n<p>(ii) Given sin<sup>-1<\/sup>(sin 7\u03c0\/6)<\/p>\n<p>But we know that sin 7\u03c0\/6 = \u2013 \u00bd<\/p>\n<p>By substituting this in sin<sup>-1<\/sup>(sin 7\u03c0\/6) we get,<\/p>\n<p>Sin<sup>-1<\/sup>\u00a0(-1\/2)<\/p>\n<p>Now let y = sin<sup>-1<\/sup>\u00a0(-1\/2)<\/p>\n<p>\u2013 Sin y = \u00bd<\/p>\n<p>\u2013 Sin (\u03c0\/6) = \u00bd<\/p>\n<p>\u2013 Sin (\u03c0\/6) = sin (- \u03c0\/6)<\/p>\n<p>The range of principal value of sin<sup>-1<\/sup>(-\u03c0\/2, \u03c0\/2) and sin (- \u03c0\/6) = \u2013 \u00bd<\/p>\n<p>Therefore sin<sup>-1<\/sup>(sin 7\u03c0\/6) = \u2013 \u03c0\/6<\/p>\n<p>(iii) Given sin<sup>-1<\/sup>(sin 5\u03c0\/6)<\/p>\n<p>We know that the value of sin 5\u03c0\/6 is \u00bd<\/p>\n<p>By substituting this value in sin<sup>-1<\/sup>(sin 5\u03c0\/6)<\/p>\n<p>We get, sin<sup>-1<\/sup>\u00a0(1\/2)<\/p>\n<p>Now let y = sin<sup>-1<\/sup>\u00a0(1\/2)<\/p>\n<p>Sin (\u03c0\/6) = \u00bd<\/p>\n<p>The range of principal value of sin<sup>-1<\/sup>(-\u03c0\/2, \u03c0\/2) and sin (\u03c0\/6) = \u00bd<\/p>\n<p>Therefore sin<sup>-1<\/sup>(sin 5\u03c0\/6) = \u03c0\/6<\/p>\n<p>(iv) Given sin<sup>-1<\/sup>(sin 13\u03c0\/7)<\/p>\n<p>Given question can be written as sin (2\u03c0 \u2013 \u03c0\/7)<\/p>\n<p>Sin (2\u03c0 \u2013 \u03c0\/7) can be written as sin (-\u03c0\/7) [since sin (2\u03c0 \u2013 \u03b8) = sin (-\u03b8)]<\/p>\n<p>By substituting these values in sin<sup>-1<\/sup>(sin 13\u03c0\/7) we get sin<sup>-1<\/sup>(sin \u2013 \u03c0\/7)<\/p>\n<p>As sin<sup>-1<\/sup>(sin x) = x with x \u2208 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>Therefore sin<sup>-1<\/sup>(sin 13\u03c0\/7) = \u2013 \u03c0\/7<\/p>\n<p>(v) Given sin<sup>-1<\/sup>(sin 17\u03c0\/8)<\/p>\n<p>Given question can be written as sin (2\u03c0 + \u03c0\/8)<\/p>\n<p>Sin (2\u03c0 + \u03c0\/8) can be written as sin (\u03c0\/8)<\/p>\n<p>By substituting these values in sin<sup>-1<\/sup>(sin 17\u03c0\/8) we get sin<sup>-1<\/sup>(sin \u03c0\/8)<\/p>\n<p>As sin<sup>-1<\/sup>(sin x) = x with x \u2208 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>Therefore sin<sup>-1<\/sup>(sin 17\u03c0\/8) = \u03c0\/8<\/p>\n<p>(vi) Given sin<sup>-1<\/sup>{(sin \u2013 17\u03c0\/8)}<\/p>\n<p>But we know that \u2013 sin \u03b8 = sin (-\u03b8)<\/p>\n<p>Therefore (sin -17\u03c0\/8) = \u2013 sin 17\u03c0\/8<\/p>\n<p>\u2013 Sin 17\u03c0\/8 = \u2013 sin (2\u03c0 + \u03c0\/8) [since sin (2\u03c0 \u2013 \u03b8) = -sin (\u03b8)]<\/p>\n<p>It can also be written as \u2013 sin (\u03c0\/8)<\/p>\n<p>\u2013 Sin (\u03c0\/8) = sin (-\u03c0\/8) [since \u2013 sin \u03b8 = sin (-\u03b8)]<\/p>\n<p>By substituting these values in sin<sup>-1<\/sup>{(sin \u2013 17\u03c0\/8)} we get,<\/p>\n<p>Sin<sup>-1<\/sup>(sin \u2013 \u03c0\/8)<\/p>\n<p>As sin<sup>-1<\/sup>(sin x) = x with x \u2208 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>Therefore sin<sup>-1<\/sup>(sin -\u03c0\/8) = \u2013 \u03c0\/8<\/p>\n<p>(vii) Given sin<sup>-1<\/sup>(sin 3)<\/p>\n<p>We know that sin<sup>-1<\/sup>(sin x) = x with x \u2208 [-\u03c0\/2, \u03c0\/2] which is approximately equal to [-1.57, 1.57]<\/p>\n<p>But here x = 3, which does not lie on the above range,<\/p>\n<p>Therefore we know that sin (\u03c0 \u2013 x) = sin (x)<\/p>\n<p>Hence sin (\u03c0 \u2013 3) = sin (3) also \u03c0 \u2013 3 \u2208 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>Sin<sup>-1<\/sup>(sin 3) = \u03c0 \u2013 3<\/p>\n<p>(viii) Given sin<sup>-1<\/sup>(sin 4)<\/p>\n<p>We know that sin<sup>-1<\/sup>(sin x) = x with x \u2208 [-\u03c0\/2, \u03c0\/2] which is approximately equal to [-1.57, 1.57]<\/p>\n<p>But here x = 4, which does not lie on the above range,<\/p>\n<p>Therefore we know that sin (\u03c0 \u2013 x) = sin (x)<\/p>\n<p>Hence sin (\u03c0 \u2013 4) = sin (4) also \u03c0 \u2013 4 \u2208 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>Sin<sup>-1<\/sup>(sin 4) = \u03c0 \u2013 4<\/p>\n<p>(ix) Given sin<sup>-1<\/sup>(sin 12)<\/p>\n<p>We know that sin<sup>-1<\/sup>(sin x) = x with x \u2208 [-\u03c0\/2, \u03c0\/2] which is approximately equal to [-1.57, 1.57]<\/p>\n<p>But here x = 12, which does not lie on the above range,<\/p>\n<p>Therefore we know that sin (2n\u03c0 \u2013 x) = sin (-x)<\/p>\n<p>Hence sin (2n\u03c0 \u2013 12) = sin (-12)<\/p>\n<p>Here n = 2 also 12 \u2013 4\u03c0 \u2208 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>Sin<sup>-1<\/sup>(sin 12) = 12 \u2013 4\u03c0<\/p>\n<p>(x) Given sin<sup>-1<\/sup>(sin 2)<\/p>\n<p>We know that sin<sup>-1<\/sup>(sin x) = x with x \u2208 [-\u03c0\/2, \u03c0\/2] which is approximately equal to [-1.57, 1.57]<\/p>\n<p>But here x = 2, which does not lie on the above range,<\/p>\n<p>Therefore we know that sin (\u03c0 \u2013 x) = sin (x)<\/p>\n<p>Hence sin (\u03c0 \u2013 2) = sin (2) also \u03c0 \u2013 2 \u2208 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>Sin<sup>-1<\/sup>(sin 2) = \u03c0 \u2013 2<\/p>\n<p><strong>2. Evaluate each of the following:<\/strong><\/p>\n<p><strong>(i) cos<sup>-1<\/sup>{cos (-\u03c0\/4)}<\/strong><\/p>\n<p><strong>(ii) cos<sup>-1<\/sup>(cos 5\u03c0\/4)<\/strong><\/p>\n<p><strong>(iii) cos<sup>-1<\/sup>(cos 4\u03c0\/3)<\/strong><\/p>\n<p><strong>(iv) cos<sup>-1<\/sup>(cos 13\u03c0\/6)<\/strong><\/p>\n<p><strong>(v) cos<sup>-1<\/sup>(cos 3)<\/strong><\/p>\n<p><strong>(vi) cos<sup>-1<\/sup>(cos 4)<\/strong><\/p>\n<p><strong>(vii) cos<sup>-1<\/sup>(cos 5)<\/strong><\/p>\n<p><strong>(viii) cos<sup>-1<\/sup>(cos 12)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given cos<sup>-1<\/sup>{cos (-\u03c0\/4)}<\/p>\n<p>We know that cos (-\u03c0\/4) = cos (\u03c0\/4) [since cos (-\u03b8) = cos \u03b8<\/p>\n<p>Also know that cos (\u03c0\/4) = 1\/\u221a2<\/p>\n<p>By substituting these values in cos<sup>-1<\/sup>{cos (-\u03c0\/4)} we get,<\/p>\n<p>Cos<sup>-1<\/sup>(1\/\u221a2)<\/p>\n<p>Now let y = cos<sup>-1<\/sup>(1\/\u221a2)<\/p>\n<p>Therefore cos y = 1\/\u221a2<\/p>\n<p>Hence range of principal value of cos<sup>-1<\/sup>\u00a0is [0, \u03c0] and cos (\u03c0\/4) = 1\/\u221a2<\/p>\n<p>Therefore cos<sup>-1<\/sup>{cos (-\u03c0\/4)} = \u03c0\/4<\/p>\n<p>(ii) Given cos<sup>-1<\/sup>(cos 5\u03c0\/4)<\/p>\n<p>But we know that cos (5\u03c0\/4) = -1\/\u221a2<\/p>\n<p>By substituting these values in cos<sup>-1<\/sup>{cos (5\u03c0\/4)} we get,<\/p>\n<p>Cos<sup>-1<\/sup>(-1\/\u221a2)<\/p>\n<p>Now let y = cos<sup>-1<\/sup>(-1\/\u221a2)<\/p>\n<p>Therefore cos y = \u2013 1\/\u221a2<\/p>\n<p>\u2013 Cos (\u03c0\/4) = 1\/\u221a2<\/p>\n<p>Cos (\u03c0 \u2013 \u03c0\/4) = \u2013 1\/\u221a2<\/p>\n<p>Cos (3 \u03c0\/4) = \u2013 1\/\u221a2<\/p>\n<p>Hence range of principal value of cos<sup>-1<\/sup>\u00a0is [0, \u03c0] and cos (3\u03c0\/4) = -1\/\u221a2<\/p>\n<p>Therefore cos<sup>-1<\/sup>{cos (5\u03c0\/4)} = 3\u03c0\/4<\/p>\n<p>(iii) Given cos<sup>-1<\/sup>(cos 4\u03c0\/3)<\/p>\n<p>But we know that cos (4\u03c0\/3) = -1\/2<\/p>\n<p>By substituting these values in cos<sup>-1<\/sup>{cos (4\u03c0\/3)} we get,<\/p>\n<p>Cos<sup>-1<\/sup>(-1\/2)<\/p>\n<p>Now let y = cos<sup>-1<\/sup>(-1\/2)<\/p>\n<p>Therefore cos y = \u2013 1\/2<\/p>\n<p>\u2013 Cos (\u03c0\/3) = 1\/2<\/p>\n<p>Cos (\u03c0 \u2013 \u03c0\/3) = \u2013 1\/2<\/p>\n<p>Cos (2\u03c0\/3) = \u2013 1\/2<\/p>\n<p>Hence range of principal value of cos<sup>-1<\/sup>\u00a0is [0, \u03c0] and cos (2\u03c0\/3) = -1\/2<\/p>\n<p>Therefore cos<sup>-1<\/sup>{cos (4\u03c0\/3)} = 2\u03c0\/3<\/p>\n<p>(iv) Given cos<sup>-1<\/sup>(cos 13\u03c0\/6)<\/p>\n<p>But we know that cos (13\u03c0\/6) = \u221a3\/2<\/p>\n<p>By substituting these values in cos<sup>-1<\/sup>{cos (13\u03c0\/6)} we get,<\/p>\n<p>Cos<sup>-1<\/sup>(\u221a3\/2)<\/p>\n<p>Now let y = cos<sup>-1<\/sup>(\u221a3\/2)<\/p>\n<p>Therefore cos y = \u221a3\/2<\/p>\n<p>Cos (\u03c0\/6) = \u221a3\/2<\/p>\n<p>Hence range of principal value of cos<sup>-1<\/sup>\u00a0is [0, \u03c0] and cos (\u03c0\/6) = \u221a3\/2<\/p>\n<p>Therefore cos<sup>-1<\/sup>{cos (13\u03c0\/6)} = \u03c0\/6<\/p>\n<p>(v) Given cos<sup>-1<\/sup>(cos 3)<\/p>\n<p>We know that cos<sup>-1<\/sup>(cos \u03b8) = \u03b8 if 0 \u2264 \u03b8 \u2264 \u03c0<\/p>\n<p>Therefore by applying this in given question we get,<\/p>\n<p>Cos<sup>-1<\/sup>(cos 3) = 3, 3 \u2208 [0, \u03c0]<\/p>\n<p>(vi) Given cos<sup>-1<\/sup>(cos 4)<\/p>\n<p>We have cos<sup>\u20131<\/sup>(cos x) = x if x\u00a0\u03f5\u00a0[0, \u03c0] \u2248 [0, 3.14]<\/p>\n<p>And here x = 4 which does not lie in the above range.<\/p>\n<p>We know that cos (2\u03c0 \u2013 x) = cos(x)<\/p>\n<p>Thus, cos (2\u03c0 \u2013 4) = cos (4) so 2\u03c0\u20134 belongs in [0, \u03c0]<\/p>\n<p>Hence cos<sup>\u20131<\/sup>(cos 4) = 2\u03c0 \u2013 4<\/p>\n<p>(vii) Given cos<sup>-1<\/sup>(cos 5)<\/p>\n<p>We have cos<sup>\u20131<\/sup>(cos x) = x if x\u00a0\u03f5\u00a0[0, \u03c0] \u2248 [0, 3.14]<\/p>\n<p>And here x = 5 which does not lie in the above range.<\/p>\n<p>We know that cos (2\u03c0 \u2013 x) = cos(x)<\/p>\n<p>Thus, cos (2\u03c0 \u2013 5) = cos (5) so 2\u03c0\u20135 belongs in [0, \u03c0]<\/p>\n<p>Hence cos<sup>\u20131<\/sup>(cos 5) = 2\u03c0 \u2013 5<\/p>\n<p>(viii) Given cos<sup>-1<\/sup>(cos 12)<\/p>\n<p>Cos<sup>\u20131<\/sup>(cos x) = x if x\u00a0\u03f5\u00a0[0, \u03c0] \u2248 [0, 3.14]<\/p>\n<p>And here x = 12 which does not lie in the above range.<\/p>\n<p>We know cos (2n\u03c0 \u2013 x) = cos (x)<\/p>\n<p>Cos (2n\u03c0 \u2013 12) = cos (12)<\/p>\n<p>Here n = 2.<\/p>\n<p>Also 4\u03c0 \u2013 12 belongs in [0, \u03c0]<\/p>\n<p>\u2234\u00a0cos<sup>\u20131<\/sup>(cos 12) = 4\u03c0 \u2013 12<\/p>\n<p><strong>3. Evaluate each of the following:<\/strong><\/p>\n<p><strong>(i) tan<sup>-1<\/sup>(tan \u03c0\/3)<\/strong><\/p>\n<p><strong>(ii) tan<sup>-1<\/sup>(tan 6\u03c0\/7)<\/strong><\/p>\n<p><strong>(iii) tan<sup>-1<\/sup>(tan 7\u03c0\/6)<\/strong><\/p>\n<p><strong>(iv) tan<sup>-1<\/sup>(tan 9\u03c0\/4)<\/strong><\/p>\n<p><strong>(v) tan<sup>-1<\/sup>(tan 1)<\/strong><\/p>\n<p><strong>(vi) tan<sup>-1<\/sup>(tan 2)<\/strong><\/p>\n<p><strong>(vii) tan<sup>-1<\/sup>(tan 4)<\/strong><\/p>\n<p><strong>(viii) tan<sup>-1<\/sup>(tan 12)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given tan<sup>-1<\/sup>(tan \u03c0\/3)<\/p>\n<p>As tan<sup>-1<\/sup>(tan x) = x if x \u03f5 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>By applying this condition in the given question we get,<\/p>\n<p>Tan<sup>-1<\/sup>(tan \u03c0\/3) = \u03c0\/3<\/p>\n<p>(ii) Given tan<sup>-1<\/sup>(tan 6\u03c0\/7)<\/p>\n<p>We know that tan 6\u03c0\/7 can be written as (\u03c0 \u2013 \u03c0\/7)<\/p>\n<p>Tan (\u03c0 \u2013 \u03c0\/7) = \u2013 tan \u03c0\/7<\/p>\n<p>We know that tan<sup>-1<\/sup>(tan x) = x if x \u03f5 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>Tan<sup>-1<\/sup>(tan 6\u03c0\/7) = \u2013 \u03c0\/7<\/p>\n<p>(iii) Given tan<sup>-1<\/sup>(tan 7\u03c0\/6)<\/p>\n<p>We know that tan 7\u03c0\/6 = 1\/\u221a3<\/p>\n<p>By substituting this value in tan<sup>-1<\/sup>(tan 7\u03c0\/6) we get,<\/p>\n<p>Tan<sup>-1<\/sup>\u00a0(1\/\u221a3)<\/p>\n<p>Now let tan<sup>-1<\/sup>\u00a0(1\/\u221a3) = y<\/p>\n<p>Tan y = 1\/\u221a3<\/p>\n<p>Tan (\u03c0\/6) = 1\/\u221a3<\/p>\n<p>The range of the principal value of tan<sup>-1<\/sup>\u00a0is (-\u03c0\/2, \u03c0\/2) and tan (\u03c0\/6) = 1\/\u221a3<\/p>\n<p>Therefore tan<sup>-1<\/sup>(tan 7\u03c0\/6) = \u03c0\/6<\/p>\n<p>(iv) Given tan<sup>-1<\/sup>(tan 9\u03c0\/4)<\/p>\n<p>We know that tan 9\u03c0\/4 = 1<\/p>\n<p>By substituting this value in tan<sup>-1<\/sup>(tan 9\u03c0\/4) we get,<\/p>\n<p>Tan<sup>-1<\/sup>\u00a0(1)<\/p>\n<p>Now let tan<sup>-1<\/sup>\u00a0(1) = y<\/p>\n<p>Tan y = 1<\/p>\n<p>Tan (\u03c0\/4) = 1<\/p>\n<p>The range of the principal value of tan<sup>-1<\/sup>\u00a0is (-\u03c0\/2, \u03c0\/2) and tan (\u03c0\/4) = 1<\/p>\n<p>Therefore tan<sup>-1<\/sup>(tan 9\u03c0\/4) = \u03c0\/4<\/p>\n<p>(v) Given tan<sup>-1<\/sup>(tan 1)<\/p>\n<p>But we have tan<sup>-1<\/sup>(tan x) = x if x \u03f5 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>By substituting this condition in given question<\/p>\n<p>Tan<sup>-1<\/sup>(tan 1) = 1<\/p>\n<p>(vi) Given tan<sup>-1<\/sup>(tan 2)<\/p>\n<p>As tan<sup>-1<\/sup>(tan x) = x if x \u03f5 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>But here x = 2 which does not belongs to above range<\/p>\n<p>We also have tan (\u03c0 \u2013 \u03b8) = \u2013tan (\u03b8)<\/p>\n<p>Therefore tan (\u03b8 \u2013 \u03c0) = tan (\u03b8)<\/p>\n<p>Tan (2 \u2013 \u03c0) = tan (2)<\/p>\n<p>Now 2 \u2013 \u03c0 is in the given range<\/p>\n<p>Hence tan<sup>\u20131<\/sup>\u00a0(tan 2) = 2 \u2013 \u03c0<\/p>\n<p>(vii) Given tan<sup>-1<\/sup>(tan 4)<\/p>\n<p>As tan<sup>-1<\/sup>(tan x) = x if x \u03f5 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>But here x = 4 which does not belongs to above range<\/p>\n<p>We also have tan (\u03c0 \u2013 \u03b8) = \u2013tan (\u03b8)<\/p>\n<p>Therefore tan (\u03b8 \u2013 \u03c0) = tan (\u03b8)<\/p>\n<p>Tan (4 \u2013 \u03c0) = tan (4)<\/p>\n<p>Now 4 \u2013 \u03c0 is in the given range<\/p>\n<p>Hence tan<sup>\u20131<\/sup>\u00a0(tan 2) = 4 \u2013 \u03c0<\/p>\n<p>(viii) Given tan<sup>-1<\/sup>(tan 12)<\/p>\n<p>As tan<sup>-1<\/sup>(tan x) = x if x \u03f5 [-\u03c0\/2, \u03c0\/2]<\/p>\n<p>But here x = 12 which does not belongs to above range<\/p>\n<p>We know that tan (2n\u03c0 \u2013 \u03b8) = \u2013tan (\u03b8)<\/p>\n<p>Tan (\u03b8 \u2013 2n\u03c0) = tan (\u03b8)<\/p>\n<p>Here n = 2<\/p>\n<p>Tan (12 \u2013 4\u03c0) = tan (12)<\/p>\n<p>Now 12 \u2013 4\u03c0 is in the given range<\/p>\n<p>\u2234\u00a0tan<sup>\u20131<\/sup>\u00a0(tan 12) = 12 \u2013 4\u03c0.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-other-exercises-of-rd-sharma-solutions-for-class-12-chapter-4-%e2%80%93-inverse-trigonometric-functions\"><\/span>Access other exercises of RD Sharma Solutions For Class 12 Chapter 4 \u2013 Inverse Trigonometric Functions<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-4-exercise-4-1\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.1<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-4-exercise-4-2\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.2<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-3\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.3<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-4\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.4<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-5\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.5<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-6\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.6<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-8\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.8<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-9\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.9<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-10\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.10<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-11\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.11<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-12\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.12<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-13\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.13<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-14\/\">RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.14<\/a><\/li>\n<\/ul>\n<p>We have provided complete details of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-12-solutions\/\"><strong>RD Sharma Solutions Class 12 Maths Chapter 4<\/strong><\/a> Exercise 4.7. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 12 Exam, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-12-chapter-4-exercise-47\"><\/span>FAQs on RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7<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-1629812420471\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-solutions-class-12-chapter-4-exercise-47\"><\/span>How many questions are there in RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 54 questions in RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629812828018\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-solutions-class-12-chapter-4-exercise-47-for-free\"><\/span>Is RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7 for free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, You can get RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7 for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629813033175\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-solutions-class-12-chapter-4-exercise-47-free-pdf\"><\/span>Where can I download RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 12 Chapter 4 Exercise 4.7:\u00a0Students can utilise the PDF RD Sharma Solutions to quickly clear their doubts and solve questions according to the most recent exam pattern. It allows students to assess their understanding of the concepts covered in each of Chapter 4&#8217;s exercises. Students can download RD Sharma Solutions Class &#8230; <a title=\"RD Sharma Solutions For Class 12 Maths Exercise 4.7 Chapter 4 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-chapter-4-exercise-4-7\/\" aria-label=\"More on RD Sharma Solutions For Class 12 Maths Exercise 4.7 Chapter 4 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":118753,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3429,73223,73664,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/118576"}],"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=118576"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/118576\/revisions"}],"predecessor-version":[{"id":119331,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/118576\/revisions\/119331"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/118753"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=118576"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=118576"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=118576"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}