{"id":62362,"date":"2023-09-12T14:11:00","date_gmt":"2023-09-12T08:41:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=62362"},"modified":"2023-11-09T11:35:07","modified_gmt":"2023-11-09T06:05:07","slug":"rd-sharma-solutions-class-11-maths-chapter-5","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/","title":{"rendered":"RD Sharma Solutions Class 11 Maths Chapter 5 &#8211; Trigonometric Functions (Updated For 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-119052\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-5.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 5 \" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-5.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-5-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 5-Trigonometric Functions:<\/strong> RD Sharma Solutions Class 11 Maths Chapter 5 is here for you to prepare the 5th chapter (Trigonometric Functions) from your class 11 Mathematics syllabus. Chapter 5 from class 11 talks about Trigonometric Functions which is one of the Important chapters which needs clarity. And with a significant guide by side, getting thorough with the chapter might make you confident with the subject.<\/p>\n<ul>\n<li style=\"text-align: left;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener noreferrer\">Class 11 Maths RD Sharma Solutions PDF<\/a><\/li>\n<\/ul>\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-69e181d1964c2\" 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: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69e181d1964c2\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#download-rd-sharma-solutions-class-11-maths-chapter-5-pdf\" title=\"Download RD Sharma Solutions Class 11 Maths Chapter 5 PDF\">Download RD Sharma Solutions Class 11 Maths Chapter 5 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#exercise-wise-rd-sharma-solutions-class-11-maths-chapter-5\" title=\"Exercise wise: RD Sharma Solutions Class 11 Maths Chapter 5\">Exercise wise: RD Sharma Solutions Class 11 Maths Chapter 5<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#access-rd-sharma-solutions-class-11-maths-chapter-5\" title=\"Access RD Sharma Solutions Class 11 Maths Chapter 5\">Access RD Sharma Solutions Class 11 Maths Chapter 5<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#exercise-51-page-no-518\" title=\"EXERCISE 5.1 PAGE NO: 5.18\">EXERCISE 5.1 PAGE NO: 5.18<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#prove-the-following-identities\" title=\"Prove the following identities:\">Prove the following identities:<\/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-11-maths-chapter-5\/#1-sec4-x-%e2%80%93-sec2-x-tan4-x-tan2-x\" title=\"1. sec4\u00a0x \u2013 sec2\u00a0x = tan4\u00a0x + tan2\u00a0x\">1. sec4\u00a0x \u2013 sec2\u00a0x = tan4\u00a0x + tan2\u00a0x<\/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-11-maths-chapter-5\/#2-sin6-x-cos6-x-1-%e2%80%93-3-sin2-x-cos2-x\" title=\"2. sin6\u00a0x + cos6\u00a0x = 1 \u2013 3 sin2\u00a0x cos2\u00a0x\">2. sin6\u00a0x + cos6\u00a0x = 1 \u2013 3 sin2\u00a0x cos2\u00a0x<\/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-solutions-class-11-maths-chapter-5\/#3-cosec-x-%e2%80%93-sin-x-sec-x-%e2%80%93-cos-x-tan-x-cot-x-1\" title=\"3. (cosec x \u2013 sin x) (sec x \u2013 cos x) (tan x + cot x) = 1\">3. (cosec x \u2013 sin x) (sec x \u2013 cos x) (tan x + cot x) = 1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#4-cosec-x-sec-x-%e2%80%93-1-%e2%80%93-cot-x-1-%e2%80%93-cos-x-tan-x-%e2%80%93-sin-x\" title=\"4. cosec x (sec x \u2013 1) \u2013 cot x (1 \u2013 cos x) = tan x \u2013 sin x\">4. cosec x (sec x \u2013 1) \u2013 cot x (1 \u2013 cos x) = tan x \u2013 sin x<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#5\" title=\"5.\u00a0\">5.\u00a0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#6\" title=\"6.\">6.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#8-sec-x-sec-y-tan-x-tan-y2-%e2%80%93-sec-x-tan-y-tan-x-sec-y2-1\" title=\"8. (sec x sec y + tan x tan y)2\u00a0\u2013 (sec x tan y + tan x sec y)2\u00a0= 1\">8. (sec x sec y + tan x tan y)2\u00a0\u2013 (sec x tan y + tan x sec y)2\u00a0= 1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#9\" title=\"9.\">9.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#10\" title=\"10.\">10.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#exercise-53-page-no-539\" title=\"EXERCISE 5.3 PAGE NO: 5.39\">EXERCISE 5.3 PAGE NO: 5.39<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-16\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#1-find-the-values-of-the-following-trigonometric-ratios\" title=\"1. Find the values of the following trigonometric ratios:\">1. Find the values of the following trigonometric ratios:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-17\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#i-sin-5%cf%803\" title=\"(i) sin 5\u03c0\/3\">(i) sin 5\u03c0\/3<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-18\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#ii-sin-17%cf%80\" title=\"(ii) sin 17\u03c0\">(ii) sin 17\u03c0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-19\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#iii-tan-11%cf%806\" title=\"(iii) tan 11\u03c0\/6\">(iii) tan 11\u03c0\/6<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-20\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#iv-cos-25%cf%804\" title=\"(iv) cos (-25\u03c0\/4)\">(iv) cos (-25\u03c0\/4)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-21\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#v-tan-7%cf%804\" title=\"(v) tan 7\u03c0\/4\">(v) tan 7\u03c0\/4<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-22\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#vi-sin-17%cf%806\" title=\"(vi) sin 17\u03c0\/6\">(vi) sin 17\u03c0\/6<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-23\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#vii-cos-19%cf%806\" title=\"(vii) cos 19\u03c0\/6\">(vii) cos 19\u03c0\/6<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-24\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#viii-sin-11%cf%806\" title=\"(viii) sin (-11\u03c0\/6)\">(viii) sin (-11\u03c0\/6)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-25\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#ix-cosec-20%cf%803\" title=\"(ix) cosec (-20\u03c0\/3)\">(ix) cosec (-20\u03c0\/3)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-26\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#x-tan-13%cf%804\" title=\"(x) tan (-13\u03c0\/4)\">(x) tan (-13\u03c0\/4)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-27\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#xi-cos-19%cf%804\" title=\"(xi) cos 19\u03c0\/4\">(xi) cos 19\u03c0\/4<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-28\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#xii-sin-41%cf%804\" title=\"(xii) sin 41\u03c0\/4\">(xii) sin 41\u03c0\/4<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-29\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#xiii-cos-39%cf%804\" title=\"(xiii) cos 39\u03c0\/4\">(xiii) cos 39\u03c0\/4<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-30\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#xiv-sin-151%cf%806\" title=\"(xiv) sin 151\u03c0\/6\">(xiv) sin 151\u03c0\/6<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-31\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#2-prove-that-i-tan-225o-cot-405o-tan-765o-cot-675o-0\" title=\"2. prove that: (i) tan 225o\u00a0cot 405o\u00a0+ tan 765o\u00a0cot 675o\u00a0= 0\">2. prove that: (i) tan 225o\u00a0cot 405o\u00a0+ tan 765o\u00a0cot 675o\u00a0= 0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-32\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#ii-sin-8%cf%803-cos-23%cf%806-cos-13%cf%803-sin-35%cf%806-12\" title=\"(ii) sin 8\u03c0\/3 cos 23\u03c0\/6 + cos 13\u03c0\/3 sin 35\u03c0\/6 = 1\/2\">(ii) sin 8\u03c0\/3 cos 23\u03c0\/6 + cos 13\u03c0\/3 sin 35\u03c0\/6 = 1\/2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-33\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#iii-cos-24o-cos-55o-cos-125o-cos-204o-cos-300o-12\" title=\"(iii) cos 24o\u00a0+ cos 55o\u00a0+ cos 125o\u00a0+ cos 204o\u00a0+ cos 300o\u00a0= 1\/2\">(iii) cos 24o\u00a0+ cos 55o\u00a0+ cos 125o\u00a0+ cos 204o\u00a0+ cos 300o\u00a0= 1\/2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-34\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#iv-tan-125o-cot-405o-%e2%80%93-tan-765o-cot-675o-0\" title=\"(iv) tan (-125o) cot (-405o) \u2013 tan (-765o) cot (675o) = 0\">(iv) tan (-125o) cot (-405o) \u2013 tan (-765o) cot (675o) = 0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-35\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#v-cos-570o-sin-510o-sin-330o-cos-390o-0\" title=\"(v) cos 570o\u00a0sin 510o\u00a0+ sin (-330o) cos (-390o) = 0\">(v) cos 570o\u00a0sin 510o\u00a0+ sin (-330o) cos (-390o) = 0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-36\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#vi-tan-11%cf%803-%e2%80%93-2-sin-4%cf%806-%e2%80%93-34-cosec2-%cf%804-4-cos2-17%cf%806-3-%e2%80%93-4%e2%88%9a32\" title=\"(vi) tan 11\u03c0\/3 \u2013 2 sin 4\u03c0\/6 \u2013 3\/4 cosec2\u00a0\u03c0\/4 + 4 cos2\u00a017\u03c0\/6 = (3 \u2013 4\u221a3)\/2\">(vi) tan 11\u03c0\/3 \u2013 2 sin 4\u03c0\/6 \u2013 3\/4 cosec2\u00a0\u03c0\/4 + 4 cos2\u00a017\u03c0\/6 = (3 \u2013 4\u221a3)\/2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-37\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#vii-3-sin-%cf%806-sec-%cf%803-%e2%80%93-4-sin-5%cf%806-cot-%cf%804-1\" title=\"(vii) 3 sin \u03c0\/6 sec \u03c0\/3 \u2013 4 sin 5\u03c0\/6 cot \u03c0\/4 = 1\">(vii) 3 sin \u03c0\/6 sec \u03c0\/3 \u2013 4 sin 5\u03c0\/6 cot \u03c0\/4 = 1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-38\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#3-prove-that\" title=\"3. Prove that:\">3. Prove that:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-39\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#i\" title=\"(i)\">(i)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-40\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#ii\" title=\"(ii)\">(ii)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-41\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#iii\" title=\"(iii)\">(iii)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-42\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#iv\" title=\"(iv)\">(iv)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-43\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#v\" title=\"(v)\">(v)<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-44\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#access-other-important-chapters-of-rd-sharma-solutions-class-11-maths\" title=\"Access Other Important Chapters of RD Sharma Solutions Class 11 Maths\u00a0\">Access Other Important Chapters of RD Sharma Solutions Class 11 Maths\u00a0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-45\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#faqs-on-rd-sharma-solutions-class-11-maths-chapter-5\" title=\"FAQs on RD Sharma Solutions Class 11 Maths Chapter 5\">FAQs on RD Sharma Solutions Class 11 Maths Chapter 5<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-46\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#from-where-can-i-download-the-rd-sharma-solutions-class-11-maths-chapter-5-pdf\" title=\"From where can I download the RD Sharma Solutions Class 11 Maths Chapter 5 PDF?\">From where can I download the RD Sharma Solutions Class 11 Maths Chapter 5 PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-47\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-5\" title=\"How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 5?\">How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 5?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-48\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/#are-the-solutions-of-rd-sharma-solutions-class-11-maths-chapter-5-reliable\" title=\"Are the solutions of RD Sharma Solutions Class 11 Maths Chapter 5 reliable?\">Are the solutions of RD Sharma Solutions Class 11 Maths Chapter 5 reliable?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-5-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 5 PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-5-1-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 5<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<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-Solutions-Class-11-Maths-Chapter-5-1-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-11-maths-chapter-5\"><\/span>Exercise wise: RD Sharma Solutions Class 11 Maths Chapter 5<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%; height: 153px;\">\n<tbody>\n<tr style=\"height: 51px;\">\n<td style=\"width: 100%; height: 51px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5-exercise-5-1\/\">Class 11 RD Sharma Chapter 5 Exercise 5.1<\/a><\/td>\n<\/tr>\n<tr style=\"height: 51px;\">\n<td style=\"width: 100%; height: 51px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5-exercise-5-2\/\">Class 11 RD Sharma Chapter 5 Exercise 5.2<\/a><\/td>\n<\/tr>\n<tr style=\"height: 51px;\">\n<td style=\"width: 100%; height: 51px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5-exercise-5-3\/\">Class 11 RD Sharma Chapter 5 Exercise 5.3<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h3><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-11-maths-chapter-5\"><\/span><span style=\"font-size: 30px; font-weight: bold; background-color: initial;\">Access RD Sharma Solutions Class 11 Maths Chapter 5<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"exercise-51-page-no-518\"><\/span>EXERCISE 5.1 PAGE NO: 5.18<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"prove-the-following-identities\"><\/span><strong>Prove the following identities:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"1-sec4-x-%e2%80%93-sec2-x-tan4-x-tan2-x\"><\/span><strong>1. sec<sup>4\u00a0<\/sup>x \u2013 sec<sup>2\u00a0<\/sup>x = tan<sup>4\u00a0<\/sup>x + tan<sup>2\u00a0<\/sup>x<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS: sec<sup>4\u00a0<\/sup>x \u2013 sec<sup>2\u00a0<\/sup>x<\/p>\n<p>(sec<sup>2\u00a0<\/sup>x)<sup>2<\/sup>\u00a0\u2013 sec<sup>2\u00a0<\/sup>x<\/p>\n<p>By using the formula, sec<sup>2<\/sup>\u00a0\u03b8 = 1 + tan<sup>2<\/sup>\u00a0\u03b8.<\/p>\n<p>(1 + tan<sup>2\u00a0<\/sup>x)\u00a0<sup>2<\/sup>\u00a0\u2013 (1 + tan<sup>2\u00a0<\/sup>x)<\/p>\n<p>1 + 2tan<sup>2\u00a0<\/sup>x + tan<sup>4\u00a0<\/sup>x \u2013 1 \u2013 tan<sup>2\u00a0<\/sup>x<\/p>\n<p>tan<sup>4\u00a0<\/sup>x + tan<sup>2\u00a0<\/sup>x<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"2-sin6-x-cos6-x-1-%e2%80%93-3-sin2-x-cos2-x\"><\/span>2. sin<sup>6\u00a0<\/sup>x + cos<sup>6\u00a0<\/sup>x = 1 \u2013 3 sin<sup>2\u00a0<\/sup>x cos<sup>2\u00a0<\/sup>x<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS: sin<sup>6\u00a0<\/sup>x + cos<sup>6\u00a0<\/sup>x<\/p>\n<p>(sin<sup>2\u00a0<\/sup>x)<sup>\u00a03<\/sup>\u00a0+ (cos<sup>2\u00a0<\/sup>x)<sup>\u00a03<\/sup><\/p>\n<p>By using the formula, a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0= (a + b) (a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0\u2013 ab)<\/p>\n<p>(sin<sup>2\u00a0<\/sup>x + cos<sup>2\u00a0<\/sup>x) [(sin<sup>2\u00a0<\/sup>x)<sup>\u00a02<\/sup>\u00a0+ (cos<sup>2\u00a0<\/sup>x)<sup>\u00a02<\/sup>\u00a0\u2013 sin<sup>2\u00a0<\/sup>x cos<sup>2\u00a0<\/sup>x]<\/p>\n<p>By using the formula, sin<sup>2\u00a0<\/sup>x + cos<sup>2\u00a0<\/sup>x = 1 and a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0= (a + b)<sup>\u00a02<\/sup>\u00a0\u2013 2ab<\/p>\n<p>1 \u00d7 [(sin<sup>2\u00a0<\/sup>x + cos<sup>2\u00a0<\/sup>x)<sup>\u00a02<\/sup>\u00a0\u2013 2sin<sup>2\u00a0<\/sup>x cos<sup>2\u00a0<\/sup>x \u2013 sin<sup>2\u00a0<\/sup>x cos<sup>2\u00a0<\/sup>x<\/p>\n<p>1<sup>2<\/sup>\u00a0\u2013 3sin<sup>2\u00a0<\/sup>x cos<sup>2\u00a0<\/sup>x<\/p>\n<p>1 \u2013 3sin<sup>2\u00a0<\/sup>x cos<sup>2\u00a0<\/sup>x<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"3-cosec-x-%e2%80%93-sin-x-sec-x-%e2%80%93-cos-x-tan-x-cot-x-1\"><\/span>3. (cosec x \u2013 sin x) (sec x \u2013 cos x) (tan x + cot x) = 1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS: (cosec x \u2013 sin x) (sec x \u2013 cos x) (tan x + cot x)<\/p>\n<p>By using the formulas<\/p>\n<p>cosec \u03b8 = 1\/sin \u03b8;<\/p>\n<p>sec \u03b8 = 1\/cos \u03b8;<\/p>\n<p>tan \u03b8 = sin \u03b8 \/ cos \u03b8;<\/p>\n<p>cot \u03b8 = cos \u03b8 \/ sin \u03b8<\/p>\n<p>Now,<\/p>\n<p><img class=\"alignnone size-full wp-image-117984\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-sharma-solutions-for-class-11-maths-chapter-5.png\" alt=\"\" width=\"355\" height=\"267\"><\/p>\n<p>1 = RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"4-cosec-x-sec-x-%e2%80%93-1-%e2%80%93-cot-x-1-%e2%80%93-cos-x-tan-x-%e2%80%93-sin-x\"><\/span>4. cosec x (sec x \u2013 1) \u2013 cot x (1 \u2013 cos x) = tan x \u2013 sin x<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS: cosec x (sec x \u2013 1) \u2013 cot x (1 \u2013 cos x)<\/p>\n<p>By using the formulas<\/p>\n<p>cosec \u03b8 = 1\/sin \u03b8;<\/p>\n<p>sec \u03b8 = 1\/cos \u03b8;<\/p>\n<p>tan \u03b8 = sin \u03b8 \/ cos \u03b8;<\/p>\n<p>cot \u03b8 = cos \u03b8 \/ sin \u03b8<\/p>\n<p>Now,<\/p>\n<p><img class=\"alignnone size-full wp-image-117993\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-51.png\" alt=\"\" width=\"282\" height=\"224\"><\/p>\n<p>By using the formula, 1 \u2013 cos<sup>2<\/sup>x = sin<sup>2<\/sup>x;<\/p>\n<p><img class=\"alignnone size-full wp-image-117997\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-52.png\" alt=\"\" width=\"170\" height=\"201\"><\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence Proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"5\"><\/span>5.\u00a0<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><img class=\"alignnone size-full wp-image-118001\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-53.png\" alt=\"\" width=\"392\" height=\"62\"><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider the LHS:<\/p>\n<p><img class=\"alignnone size-full wp-image-118003\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-54.png\" alt=\"\" width=\"328\" height=\"62\"><\/p>\n<p>By using the formula,<\/p>\n<p>cosec \u03b8 = 1\/sin \u03b8;<\/p>\n<p>sec \u03b8 = 1\/cos \u03b8;<\/p>\n<p>Now,<\/p>\n<p><img class=\"alignnone size-full wp-image-118004\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-55.png\" alt=\"\" width=\"556\" height=\"413\"><\/p>\n<p><img class=\"alignnone size-full wp-image-118005\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-56.png\" alt=\"\" width=\"556\" height=\"413\"><\/p>\n<p>sin x<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence Proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"6\"><\/span><strong>6.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><img class=\"alignnone size-full wp-image-118009\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-5.gif\" alt=\"\" width=\"329\" height=\"37\"><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider the LHS:<\/p>\n<p><img class=\"alignnone size-full wp-image-118012\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-57.png\" alt=\"\" width=\"172\" height=\"53\"><\/p>\n<p>By using the formula,<\/p>\n<p>tan \u03b8 = sin \u03b8 \/ cos \u03b8;<\/p>\n<p>cot \u03b8 = cos \u03b8 \/ sin \u03b8<\/p>\n<p>Now,<\/p>\n<p><img class=\"alignnone size-full wp-image-118031\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-58.png\" alt=\"\" width=\"336\" height=\"331\"><\/p>\n<p>By using the formula, a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0= (a \u2013 b) (a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0+ ab)<\/p>\n<p><img class=\"alignnone size-full wp-image-118032\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-59.png\" alt=\"\" width=\"373\" height=\"280\"><\/p>\n<p>By using the formula,<\/p>\n<p>cosec \u03b8 = 1\/sin \u03b8,<\/p>\n<p>sec \u03b8 = 1\/cos \u03b8;<\/p>\n<p>cosec x \u00d7 sec x + 1<\/p>\n<p>sec x cosec x + 1<\/p>\n<p>=RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>7.\u00a0<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-118035\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-510.png\" alt=\"\" width=\"313\" height=\"72\"><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p><img class=\"alignnone size-full wp-image-118036\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-511.png\" alt=\"\" width=\"276\" height=\"67\"><\/p>\n<p>By using the formula a<sup>3<\/sup>\u00a0\u00b1 b<sup>3<\/sup>\u00a0= (a \u00b1 b) (a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u2213\u00a0ab)<\/p>\n<p><img class=\"alignnone size-full wp-image-118038\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-512.png\" alt=\"\" width=\"468\" height=\"100\"><\/p>\n<p>We know, sin<sup>2<\/sup>x + cos<sup>2<\/sup>x = 1.<\/p>\n<p>1 \u2013 sinx\u00a0cosx\u00a0+ 1 + sinx cosx<\/p>\n<p>2<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence Proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"8-sec-x-sec-y-tan-x-tan-y2-%e2%80%93-sec-x-tan-y-tan-x-sec-y2-1\"><\/span>8. (sec x sec y + tan x tan y)<sup>2<\/sup>\u00a0\u2013 (sec x tan y + tan x sec y)<sup>2<\/sup>\u00a0= 1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>(sec x sec y + tan x tan y)<sup>2<\/sup>\u00a0\u2013 (sec x tan y + tan x sec y)<sup>2<\/sup><\/p>\n<p>Expanding the above equation we get,<\/p>\n<p>[(sec x sec y)<sup>2<\/sup>\u00a0+ (tan x tan y)<sup>2<\/sup>\u00a0+ 2 (sec x sec y) (tan x tan y)] \u2013 [(sec x tan y)<sup>2<\/sup>\u00a0+ (tan x sec y)<sup>2<\/sup>\u00a0+ 2 (sec x tan y) (tan x sec y)] [sec<sup>2\u00a0<\/sup>x sec<sup>2<\/sup>\u00a0y + tan<sup>2\u00a0<\/sup>x tan<sup>2<\/sup>\u00a0y + 2 (sec x sec y) (tan x tan y)] \u2013 [sec<sup>2\u00a0<\/sup>x tan<sup>2<\/sup>\u00a0y + tan<sup>2\u00a0<\/sup>x sec<sup>2<\/sup>\u00a0y + 2 (sec<sup>2\u00a0<\/sup>x tan<sup>2<\/sup>\u00a0y) (tan x sec y)]<\/p>\n<p>sec<sup>2\u00a0<\/sup>x sec<sup>2<\/sup>\u00a0y \u2013 sec<sup>2\u00a0<\/sup>x tan<sup>2<\/sup>\u00a0y + tan<sup>2\u00a0<\/sup>x tan<sup>2<\/sup>\u00a0y \u2013 tan<sup>2\u00a0<\/sup>x sec<sup>2<\/sup>\u00a0y<\/p>\n<p>sec<sup>2\u00a0<\/sup>x (sec<sup>2<\/sup>\u00a0y \u2013 tan<sup>2<\/sup>\u00a0y) + tan<sup>2\u00a0<\/sup>x (tan<sup>2<\/sup>\u00a0y \u2013 sec<sup>2<\/sup>\u00a0y)<\/p>\n<p>sec<sup>2\u00a0<\/sup>x (sec<sup>2<\/sup>\u00a0y \u2013 tan<sup>2<\/sup>\u00a0y) \u2013 tan<sup>2\u00a0<\/sup>x (sec<sup>2<\/sup>\u00a0y \u2013 tan<sup>2<\/sup>\u00a0y)<\/p>\n<p>We know, sec<sup>2\u00a0<\/sup>x \u2013 tan<sup>2\u00a0<\/sup>x = 1.<\/p>\n<p>sec<sup>2\u00a0<\/sup>x \u00d7 1 \u2013 tan<sup>2\u00a0<\/sup>x \u00d7 1<\/p>\n<p>sec<sup>2\u00a0<\/sup>x \u2013 tan<sup>2\u00a0<\/sup>x<\/p>\n<p>1 = RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"9\"><\/span><strong>9.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><img class=\"alignnone size-full wp-image-118042\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-513.png\" alt=\"\" width=\"259\" height=\"63\"><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us Consider RHS:<\/p>\n<p><img class=\"alignnone size-full wp-image-118043\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-514.png\" alt=\"\" width=\"340\" height=\"166\"><\/p>\n<p><img class=\"alignnone size-full wp-image-118061\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-515-1.png\" alt=\"\" width=\"410\" height=\"525\"><\/p>\n<p><img class=\"alignnone size-full wp-image-118045\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rds-516.png\" alt=\"\" width=\"288\" height=\"313\"><\/p>\n<p><img class=\"alignnone size-full wp-image-118047\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-517.png\" alt=\"\" width=\"223\" height=\"310\"><\/p>\n<p>= LHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence Proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"10\"><\/span><strong>10.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><img class=\"alignnone size-full wp-image-118051\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-518.png\" alt=\"\" width=\"377\" height=\"67\"><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p><img class=\"alignnone size-full wp-image-118053\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-518-1.png\" alt=\"\" width=\"377\" height=\"67\"><\/p>\n<p>By using the formulas,<\/p>\n<p>1 + tan<sup>2<\/sup>x = sec<sup>2<\/sup>x and 1 + cot<sup>2<\/sup>x = cosec<sup>2<\/sup>x<\/p>\n<p><img class=\"alignnone size-full wp-image-118056\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-520.png\" alt=\"\" width=\"140\" height=\"186\"><\/p>\n<p><img class=\"alignnone size-full wp-image-118064\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-521.png\" alt=\"\" width=\"281\" height=\"316\"><\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>11<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-118068\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-5-1.gif\" alt=\"\" width=\"297\" height=\"43\"><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p><img class=\"alignnone size-full wp-image-118069\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-522.png\" alt=\"\" width=\"208\" height=\"59\"><\/p>\n<p>By using the formula,<\/p>\n<p>tan \u03b8 = sin \u03b8 \/ cos \u03b8;<\/p>\n<p>cot \u03b8 = cos \u03b8 \/ sin \u03b8<\/p>\n<p>Now,<\/p>\n<p><img class=\"alignnone size-full wp-image-118072\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-523.png\" alt=\"\" width=\"249\" height=\"194\"><\/p>\n<p>By using the formula, a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0= (a + b) (a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u2013\u00a0ab)<\/p>\n<p><img class=\"alignnone size-full wp-image-118075\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-524.png\" alt=\"\" width=\"493\" height=\"143\"><\/p>\n<p>We know, sin<sup>2\u00a0<\/sup>x + cos<sup>2\u00a0<\/sup>x = 1.<\/p>\n<p>1 \u2013 1 + sin x cos x<\/p>\n<p>Sin x cos x<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>12.\u00a0<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-118079\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-5-2.gif\" alt=\"\" width=\"527\" height=\"47\"><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p><img class=\"alignnone size-full wp-image-118081\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-525.png\" alt=\"\" width=\"390\" height=\"64\"><\/p>\n<p>By using the formula,<\/p>\n<p>cosec \u03b8 = 1\/sin \u03b8,<\/p>\n<p>sec \u03b8 = 1\/cos \u03b8;<\/p>\n<p><img class=\"alignnone size-full wp-image-118087\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-526.png\" alt=\"\" width=\"427\" height=\"238\"><\/p>\n<p><img class=\"alignnone size-full wp-image-118093\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-527.png\" alt=\"\" width=\"490\" height=\"287\"><\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>13. (1 + tan \u03b1 tan \u03b2)<sup>\u00a02<\/sup>\u00a0+ (tan \u03b1 \u2013 tan \u03b2)<sup>\u00a02<\/sup>\u00a0= sec<sup>2<\/sup>\u00a0\u03b1 sec<sup>2<\/sup>\u00a0\u03b2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS: (1 + tan \u03b1 tan \u03b2)<sup>\u00a02<\/sup>\u00a0+ (tan \u03b1 \u2013 tan \u03b2)<sup>\u00a02<\/sup><\/p>\n<p>1+ tan<sup>2<\/sup>\u00a0\u03b1 tan<sup>2<\/sup>\u00a0\u03b2 + 2 tan \u03b1 tan \u03b2 + tan<sup>2<\/sup>\u00a0\u03b1 + tan<sup>2<\/sup>\u00a0\u03b2 \u2013 2 tan \u03b1 tan \u03b2<\/p>\n<p>1 + tan<sup>2<\/sup>\u00a0\u03b1 tan<sup>2<\/sup>\u00a0\u03b2 + tan<sup>2<\/sup>\u00a0\u03b1 + tan<sup>2<\/sup>\u00a0\u03b2<\/p>\n<p>tan<sup>2<\/sup>\u00a0\u03b1 (tan<sup>2<\/sup>\u00a0\u03b2 + 1) + 1 (1 + tan<sup>2<\/sup>\u00a0\u03b2)<\/p>\n<p>(1 + tan<sup>2<\/sup>\u00a0\u03b2) (1 + tan<sup>2<\/sup>\u00a0\u03b1)<\/p>\n<p>We know, 1 + tan<sup>2<\/sup>\u00a0\u03b8 = sec<sup>2<\/sup>\u00a0\u03b8<\/p>\n<p>So,<\/p>\n<p>sec<sup>2<\/sup>\u00a0\u03b1 sec<sup>2<\/sup>\u00a0\u03b2<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p>EXERCISE 5.2 PAGE NO: 5.25<\/p>\n<p><strong>1. Find the values of the other five trigonometric functions in each of the following:<\/strong><\/p>\n<p><strong>(i) cot x = 12\/5, x in quadrant III<\/strong><\/p>\n<p><strong>(ii) cos x = -1\/2, x in quadrant II<\/strong><\/p>\n<p><strong>(iii) tan x = 3\/4, x in quadrant III<\/strong><\/p>\n<p><strong>(iv) sin x = 3\/5, x in quadrant I<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>cot x = 12\/5, x in quadrant III<\/p>\n<p>In the third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative.<\/p>\n<p>By using the formulas,<\/p>\n<p>tan x = 1\/cot x<\/p>\n<p>= 1\/(12\/5)<\/p>\n<p>= 5\/12<\/p>\n<p>cosec x = \u2013<strong>\u221a<\/strong>(1 + cot<sup>2<\/sup>\u00a0x)<\/p>\n<p>= \u2013<strong>\u221a<\/strong>(1 + (12\/5)<sup>2<\/sup>)<\/p>\n<p>= \u2013<strong>\u221a<\/strong>(25+144)\/25<\/p>\n<p>= \u2013<strong>\u221a<\/strong>(169\/25)<\/p>\n<p>= -13\/5<\/p>\n<p>sin x = 1\/cosec x<\/p>\n<p>= 1\/(-13\/5)<\/p>\n<p>= -5\/13<\/p>\n<p>cos x = \u2013\u00a0<strong>\u221a<\/strong>(1 \u2013 sin<sup>2<\/sup>\u00a0x)<\/p>\n<p>= \u2013\u00a0<strong>\u221a<\/strong>(1 \u2013 (-5\/13)<sup>2<\/sup>)<\/p>\n<p>= \u2013 \u221a(169-25)\/169<\/p>\n<p>= \u2013 \u221a(144\/169)<\/p>\n<p>= -12\/13<\/p>\n<p>sec x = 1\/cos x<\/p>\n<p>= 1\/(-12\/13)<\/p>\n<p>= -13\/12<\/p>\n<p>\u2234 sin x = -5\/13, cos x = -12\/13, tan x = 5\/12, cosec x = -13\/5, sec x = -13\/12<\/p>\n<p><strong>(ii)\u00a0<\/strong>cos x = -1\/2, x in quadrant II<\/p>\n<p>In second quadrant, sin x and cosec x are positive. tan x, cot x, cos x, sec x are negative.<\/p>\n<p>By using the formulas,<\/p>\n<p>sin x = \u221a(1 \u2013 cos<sup>2<\/sup>\u00a0x)<\/p>\n<p>= \u221a(1 \u2013 (-1\/2)<sup>2<\/sup>)<\/p>\n<p>= \u221a(4-1)\/4<\/p>\n<p>= \u221a(3\/4)<\/p>\n<p>= \u221a3\/2<\/p>\n<p>tan x = sin x\/cos x<\/p>\n<p>= (\u221a3\/2)\/(-1\/2)<\/p>\n<p>= -\u221a3<\/p>\n<p>cot x = 1\/tan x<\/p>\n<p>= 1\/-\u221a3<\/p>\n<p>= -1\/\u221a3<\/p>\n<p>cosec x = 1\/sin x<\/p>\n<p>= 1\/(\u221a3\/2)<\/p>\n<p>= 2\/\u221a3<\/p>\n<p>sec x = 1\/cos x<\/p>\n<p>= 1\/(-1\/2)<\/p>\n<p>= -2<\/p>\n<p>\u2234 sin x = \u221a3\/2, tan x = -\u221a3, cosec x = 2\/\u221a3, cot x = -1\/\u221a3 sec x = -2<\/p>\n<p><strong>(iii)<\/strong>\u00a0tan x = 3\/4, x in quadrant III<\/p>\n<p>In the third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative.<\/p>\n<p>By using the formulas,<\/p>\n<p>sin x = \u221a(1 \u2013 cos<sup>2<\/sup>\u00a0x)<\/p>\n<p>= \u2013 \u221a(1-(-4\/5)<sup>2<\/sup>)<\/p>\n<p>= \u2013 \u221a(25-16)\/25<\/p>\n<p>= \u2013 \u221a(9\/25)<\/p>\n<p>= \u2013 3\/5<\/p>\n<p>cos x = 1\/sec x<\/p>\n<p>= 1\/(-5\/4)<\/p>\n<p>= -4\/5<\/p>\n<p>cot x = 1\/tan x<\/p>\n<p>= 1\/(3\/4)<\/p>\n<p>= 4\/3<\/p>\n<p>cosec x = 1\/sin x<\/p>\n<p>= 1\/(-3\/5)<\/p>\n<p>= -5\/3<\/p>\n<p>sec x = -\u221a(1 + tan<sup>2<\/sup>\u00a0x)<\/p>\n<p>= \u2013 \u221a(1+(3\/4)<sup>2<\/sup>)<\/p>\n<p>= \u2013 \u221a(16+9)\/16<\/p>\n<p>= \u2013 \u221a (25\/16)<\/p>\n<p>= -5\/4<\/p>\n<p>\u2234 sin x = -3\/5, cos x = -4\/5, cosec x = -5\/3, sec x = -5\/4, cot x = 4\/3<\/p>\n<p><strong>(iv)\u00a0<\/strong>sin x = 3\/5, x in quadrant I<\/p>\n<p>In the first quadrant, all trigonometric ratios are positive.<\/p>\n<p>So, by using the formulas,<\/p>\n<p>tan x = sin x\/cos x<\/p>\n<p>= (3\/5)\/(4\/5)<\/p>\n<p>= 3\/4<\/p>\n<p>cosec x = 1\/sin x<\/p>\n<p>= 1\/(3\/5)<\/p>\n<p>= 5\/3<\/p>\n<p>cos x = \u221a(1-sin<sup>2<\/sup>\u00a0x)<\/p>\n<p>= \u221a(1 \u2013 (-3\/5)<sup>2<\/sup>)<\/p>\n<p>= \u221a(25-9)\/25<\/p>\n<p>= \u221a(16\/25)<\/p>\n<p>= 4\/5<\/p>\n<p>sec x = 1\/cos x<\/p>\n<p>= 1\/(4\/5)<\/p>\n<p>= 5\/4<\/p>\n<p>cot x = 1\/tan x<\/p>\n<p>= 1\/(3\/4)<\/p>\n<p>= 4\/3<\/p>\n<p>\u2234 cos x = 4\/5, tan x = 3\/4, cosec x = 5\/3, sec x = 5\/4, cot x = 4\/3<\/p>\n<p><strong>2. If sin x = 12\/13\u00a0and lies in the second quadrant, find the value of sec x + tan x.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Sin x = 12\/13 and x lies in the second quadrant.<\/p>\n<p>We know, in the second quadrant, sin x, and cosec x are positive and all other ratios are negative.<\/p>\n<p>By using the formulas,<\/p>\n<p>Cos x = \u221a(1-sin<sup>2<\/sup>\u00a0x)<\/p>\n<p>= \u2013 \u221a(1-(12\/13)<sup>2<\/sup>)<\/p>\n<p>= \u2013 \u221a(1- (144\/169))<\/p>\n<p>= \u2013 \u221a(169-144)\/169<\/p>\n<p>= -\u221a(25\/169)<\/p>\n<p>= \u2013 5\/13<\/p>\n<p>We know,<\/p>\n<p>tan x = sin x\/cos x<\/p>\n<p>sec x = 1\/cos x<\/p>\n<p>Now,<\/p>\n<p>tan x = (12\/13)\/(-5\/13)<\/p>\n<p>= -12\/5<\/p>\n<p>sec x = 1\/(-5\/13)<\/p>\n<p>= -13\/5<\/p>\n<p>Sec x + tan x = -13\/5 + (-12\/5)<\/p>\n<p>= (-13-12)\/5<\/p>\n<p>= -25\/5<\/p>\n<p>= -5<\/p>\n<p>\u2234 Sec x + tan x = -5<\/p>\n<p><strong>3. If sin x = 3\/5,\u00a0tan\u00a0y = 1\/2\u00a0and \u03c0\/2 &lt; x&lt; \u03c0&lt; y&lt; 3\u03c0\/2\u00a0find the value of 8 tan x -\u221a5 sec y.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>sin x = 3\/5, tan y = 1\/2 and\u00a0\u03c0\/2 &lt; x&lt; \u03c0&lt; y&lt; 3\u03c0\/2<\/p>\n<p>We know that x is in the second quadrant and y is in the third quadrant.<\/p>\n<p>In the second quadrant, cos x and tan x are negative.<\/p>\n<p>In the third quadrant, sec y is negative.<\/p>\n<p>By using the formula,<\/p>\n<p>cos x = \u2013 \u221a(1-sin<sup>2<\/sup>\u00a0x)<\/p>\n<p>tan x = sin x\/cos x<\/p>\n<p>Now,<\/p>\n<p>cos x = \u2013 \u221a(1-sin<sup>2<\/sup>\u00a0x)<\/p>\n<p>= \u2013 \u221a(1 \u2013 (3\/5)<sup>2<\/sup>)<\/p>\n<p>= \u2013 \u221a(1 \u2013 9\/25)<\/p>\n<p>= \u2013 \u221a((25-9)\/25)<\/p>\n<p>= \u2013 \u221a(16\/25)<\/p>\n<p>= \u2013 4\/5<\/p>\n<p>tan x = sin x\/cos x<\/p>\n<p>= (3\/5)\/(-4\/5)<\/p>\n<p>= 3\/5 \u00d7 -5\/4<\/p>\n<p>= -3\/4<\/p>\n<p>We know that sec y = \u2013 \u221a(1+tan<sup>2<\/sup>\u00a0y)<\/p>\n<p>= \u2013 \u221a(1 + (1\/2)<sup>2<\/sup>)<\/p>\n<p>= \u2013 \u221a(1 + 1\/4)<\/p>\n<p>= \u2013 \u221a((4+1)\/4)<\/p>\n<p>= \u2013 \u221a(5\/4)<\/p>\n<p>= \u2013 \u221a5\/2<\/p>\n<p>Now, 8 tan x \u2013 \u221a5 sec y = 8(-3\/4) \u2013 \u221a5(-\u221a5\/2)<\/p>\n<p>= -6 + 5\/2<\/p>\n<p>= (-12+5)\/2<\/p>\n<p>= -7\/2<\/p>\n<p>\u2234 8 tan x \u2013 \u221a5 sec y = -7\/2<\/p>\n<p><strong>4. If sin x + cos x = 0 and x lies in the fourth quadrant, find sin x and cos x.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Sin x + cos x = 0 and x lies in the fourth quadrant.<\/p>\n<p>Sin x\u00a0= -cos x<\/p>\n<p>Sin x\/cos x = -1<\/p>\n<p>So, tan x = -1 (since, tan x = sin x\/cos x)<\/p>\n<p>We know that, in the fourth quadrant, cos x and sec x are positive and all other ratios are negative.<\/p>\n<p>By using the formulas,<\/p>\n<p>Sec x = \u221a(1 + tan<sup>2<\/sup>\u00a0x)<\/p>\n<p>Cos x = 1\/sec x<\/p>\n<p>Sin x = \u2013 \u221a(1- cos<sup>2<\/sup>\u00a0x)<\/p>\n<p>Now,<\/p>\n<p>Sec x = \u221a(1 + tan<sup>2<\/sup>\u00a0x)<\/p>\n<p>= \u221a(1 + (-1)<sup>2<\/sup>)<\/p>\n<p>= \u221a2<\/p>\n<p>Cos x = 1\/sec x<\/p>\n<p>= 1\/\u221a2<\/p>\n<p>Sin x = \u2013 \u221a(1 \u2013 cos<sup>2<\/sup>\u00a0x)<\/p>\n<p>= \u2013 \u221a(1 \u2013 (1\/\u221a2)<sup>2<\/sup>)<\/p>\n<p>= \u2013 \u221a(1 \u2013 (1\/2))<\/p>\n<p>= \u2013 \u221a((2-1)\/2)<\/p>\n<p>= \u2013 \u221a(1\/2)<\/p>\n<p>= -1\/\u221a2<\/p>\n<p>\u2234 sin x = -1\/\u221a2 and cos x = 1\/\u221a2<\/p>\n<p><strong>5. If cos x = -3\/5 and \u03c0&lt;x&lt;3\u03c0\/2 find the values of other five trigonometric functions and hence evaluate\u00a0<img title=\"RD Sharma Solutions for Class 11 Maths Chapter 5 \u2013 Trigonometric Functions image - 32\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-5-3.gif\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 5 \u2013 Trigonometric Functions image - 32\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>cos x= -3\/5 and \u03c0 &lt;x &lt; 3\u03c0\/2<\/p>\n<p>We know that in the third quadrant, tan x and cot x are positive and all other rations are negative.<\/p>\n<p>By using the formulas,<\/p>\n<p>Sin x = \u2013 \u221a(1-cos<sup>2<\/sup>\u00a0x)<\/p>\n<p>Tan x = sin x\/cos x<\/p>\n<p>Cot x = 1\/tan x<\/p>\n<p>Sec x = 1\/cos x<\/p>\n<p>Cosec x = 1\/sin x<\/p>\n<p>Now,<\/p>\n<p>Sin x = \u2013 \u221a(1-cos<sup>2<\/sup>\u00a0x)<\/p>\n<p>= \u2013 \u221a(1-(-3\/5)<sup>2<\/sup>)<\/p>\n<p>= \u2013 \u221a(1-9\/25)<\/p>\n<p>= \u2013 \u221a((25-9)\/25)<\/p>\n<p>= \u2013 \u221a(16\/25)<\/p>\n<p>= \u2013 4\/5<\/p>\n<p>Tan x = sin x\/cos x<\/p>\n<p>= (-4\/5)\/(-3\/5)<\/p>\n<p>= -4\/5 \u00d7 -5\/3<\/p>\n<p>= 4\/3<\/p>\n<p>Cot x = 1\/tan x<\/p>\n<p>= 1\/(4\/3)<\/p>\n<p>= 3\/4<\/p>\n<p>Sec x = 1\/cos x<\/p>\n<p>= 1\/(-3\/5)<\/p>\n<p>= -5\/3<\/p>\n<p>Cosec x = 1\/sin x<\/p>\n<p>= 1\/(-4\/5)<\/p>\n<p>= -5\/4<\/p>\n<p>\u2234\u00a0<strong><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 5 \u2013 Trigonometric Functions image - 33\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-5-4.gif\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 5 \u2013 Trigonometric Functions image - 33\" \/><\/strong><br \/>= [(-5\/4) + (3\/4)] \/ [(-5\/3) \u2013 (4\/3)]<\/p>\n<p>= [(-5+3)\/4] \/ [(-5-4)\/3]<\/p>\n<p>= [-2\/4] \/ [-9\/3]<\/p>\n<p>= [-1\/2] \/ [-3]<\/p>\n<p>= 1\/6<\/p>\n<h3><span class=\"ez-toc-section\" id=\"exercise-53-page-no-539\"><\/span>EXERCISE 5.3 PAGE NO: 5.39<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"1-find-the-values-of-the-following-trigonometric-ratios\"><\/span><strong>1. Find the values of the following trigonometric ratios:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"i-sin-5%cf%803\"><\/span>(i) sin 5\u03c0\/3<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"ii-sin-17%cf%80\"><\/span>(ii) sin 17\u03c0<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iii-tan-11%cf%806\"><\/span>(iii) tan 11\u03c0\/6<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iv-cos-25%cf%804\"><\/span>(iv) cos (-25\u03c0\/4)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"v-tan-7%cf%804\"><\/span>(v) tan 7\u03c0\/4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"vi-sin-17%cf%806\"><\/span>(vi) sin 17\u03c0\/6<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"vii-cos-19%cf%806\"><\/span>(vii) cos 19\u03c0\/6<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"viii-sin-11%cf%806\"><\/span>(viii) sin (-11\u03c0\/6)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"ix-cosec-20%cf%803\"><\/span>(ix) cosec (-20\u03c0\/3)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"x-tan-13%cf%804\"><\/span>(x) tan (-13\u03c0\/4)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"xi-cos-19%cf%804\"><\/span>(xi) cos 19\u03c0\/4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"xii-sin-41%cf%804\"><\/span>(xii) sin 41\u03c0\/4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"xiii-cos-39%cf%804\"><\/span>(xiii) cos 39\u03c0\/4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"xiv-sin-151%cf%806\"><\/span>(xiv) sin 151\u03c0\/6<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>sin 5\u03c0\/3<\/p>\n<p>5\u03c0\/3 = (5\u03c0\/3 \u00d7 180)<sup>o<\/sup><\/p>\n<p>= 300<sup>o<\/sup><\/p>\n<p>= (90\u00d73 + 30)<sup>o<\/sup><\/p>\n<p>Since, 300<sup>o<\/sup> lies in the IV quadrant in which the sine function is negative.<\/p>\n<p>sin 5\u03c0\/3 = sin (300)<sup>o<\/sup><\/p>\n<p>= sin (90\u00d73 + 30)<sup>o<\/sup><\/p>\n<p>= \u2013 cos 30<sup>o<\/sup><\/p>\n<p>= \u2013 \u221a3\/2<\/p>\n<p><strong>(ii)\u00a0<\/strong>sin 17\u03c0<\/p>\n<p>Sin 17\u03c0 = sin 3060<sup>o<\/sup><\/p>\n<p>= sin (90\u00d734 + 0)<sup>o<\/sup><\/p>\n<p>Since, 3060<sup>o<\/sup> lies in the negative direction of the x-axis i.e., on the boundary line of II and III quadrants.<\/p>\n<p>Sin 17\u03c0 = sin (90\u00d734 + 0)<sup>o<\/sup><\/p>\n<p>= \u2013 sin 0<sup>o<\/sup><\/p>\n<p>= 0<\/p>\n<p><strong>(iii)<\/strong>\u00a0tan 11\u03c0\/6<\/p>\n<p>tan 11\u03c0\/6 = (11\/6 \u00d7 180)<sup>o<\/sup><\/p>\n<p>= 330<sup>o<\/sup><\/p>\n<p>Since, 330<sup>o<\/sup> lies in the IV quadrant in which the tangent function is negative.<\/p>\n<p>tan 11\u03c0\/6 = tan (300)<sup>o<\/sup><\/p>\n<p>= tan (90\u00d73 + 60)<sup>o<\/sup><\/p>\n<p>= \u2013 cot 60<sup>o<\/sup><\/p>\n<p>= \u2013 1\/\u221a3<\/p>\n<p><strong>(iv)<\/strong>\u00a0cos (-25\u03c0\/4)<\/p>\n<p>cos (-25\u03c0\/4) = cos (-1125)<sup>o<\/sup><\/p>\n<p>= cos (1125)<sup>o<\/sup><\/p>\n<p>Since, 1125<sup>o<\/sup> lies in the I quadrant in which the cosine function is positive.<\/p>\n<p>cos (1125)<sup>o<\/sup>\u00a0= cos (90\u00d712 + 45)<sup>o<\/sup><\/p>\n<p>= cos 45<sup>o<\/sup><\/p>\n<p>= 1\/\u221a2<\/p>\n<p><strong>(v)<\/strong>\u00a0tan 7\u03c0\/4<\/p>\n<p>tan 7\u03c0\/4 = tan 315<sup>o<\/sup><\/p>\n<p>= tan (90\u00d73 + 45)<sup>o<\/sup><\/p>\n<p>Since, 315<sup>o<\/sup> lies in the IV quadrant in which the tangent function is negative.<\/p>\n<p>tan 315<sup>o<\/sup>\u00a0= tan (90\u00d73 + 45)<sup>o<\/sup><\/p>\n<p>= \u2013 cot 45<sup>o<\/sup><\/p>\n<p>= -1<\/p>\n<p><strong>(vi)<\/strong>\u00a0sin 17\u03c0\/6<\/p>\n<p>sin 17\u03c0\/6 = sin 510<sup>o<\/sup><\/p>\n<p>= sin (90\u00d75 + 60)<sup>o<\/sup><\/p>\n<p>Since, 510<sup>o<\/sup> lies in the II quadrant in which the sine function is positive.<\/p>\n<p>sin 510<sup>o<\/sup>\u00a0= sin (90\u00d75 + 60)<sup>o<\/sup><\/p>\n<p>= cos 60<sup>o<\/sup><\/p>\n<p>= 1\/2<\/p>\n<p><strong>(vii)<\/strong>\u00a0cos 19\u03c0\/6<\/p>\n<p>cos 19\u03c0\/6 = cos 570<sup>o<\/sup><\/p>\n<p>= cos (90\u00d76 + 30)<sup>o<\/sup><\/p>\n<p>Since, 570<sup>o<\/sup> lies in the III quadrant in which the cosine function is negative.<\/p>\n<p>cos 570<sup>o<\/sup>\u00a0= cos (90\u00d76 + 30)<sup>o<\/sup><\/p>\n<p>= \u2013 cos 30<sup>o<\/sup><\/p>\n<p>= \u2013 \u221a3\/2<\/p>\n<p><strong>(viii)<\/strong>\u00a0sin (-11\u03c0\/6)<\/p>\n<p>sin (-11\u03c0\/6) = sin (-330<sup>o<\/sup>)<\/p>\n<p>= \u2013 sin (90\u00d73 + 60)<sup>o<\/sup><\/p>\n<p>Since, 330<sup>o<\/sup>\u00a0lies in the IV quadrant in which the sine function is negative.<\/p>\n<p>sin (-330<sup>o<\/sup>) = \u2013 sin (90\u00d73 + 60)<sup>o<\/sup><\/p>\n<p>= \u2013 (-cos 60<sup>o<\/sup>)<\/p>\n<p>= \u2013 (-1\/2)<\/p>\n<p>= 1\/2<\/p>\n<p><strong>(ix)<\/strong>\u00a0cosec (-20\u03c0\/3)<\/p>\n<p>cosec (-20\u03c0\/3) = cosec (-1200)<sup>o<\/sup><\/p>\n<p>= \u2013 cosec (1200)<sup>o<\/sup><\/p>\n<p>= \u2013 cosec (90\u00d713 + 30)<sup>o<\/sup><\/p>\n<p>Since, 1200<sup>o<\/sup> lies in the II quadrant in which the cosec function is positive.<\/p>\n<p>cosec (-1200)<sup>o<\/sup>\u00a0= \u2013 cosec (90\u00d713 + 30)<sup>o<\/sup><\/p>\n<p>= \u2013 sec 30<sup>o<\/sup><\/p>\n<p>= -2\/\u221a3<\/p>\n<p><strong>(x)<\/strong>\u00a0tan (-13\u03c0\/4)<\/p>\n<p>tan (-13\u03c0\/4) = tan (-585)<sup>o<\/sup><\/p>\n<p>= \u2013 tan (90\u00d76 + 45)<sup>o<\/sup><\/p>\n<p>Since, 585<sup>o<\/sup>\u00a0lies in the III quadrant in which the tangent function is positive.<\/p>\n<p>tan (-585)<sup>o<\/sup>\u00a0= \u2013 tan (90\u00d76 + 45)<sup>o<\/sup><\/p>\n<p>= \u2013 tan 45<sup>o<\/sup><\/p>\n<p>= -1<\/p>\n<p><strong>(xi)<\/strong>\u00a0cos 19\u03c0\/4<\/p>\n<p>cos 19\u03c0\/4 = cos 855<sup>o<\/sup><\/p>\n<p>= cos (90\u00d79 + 45)<sup>o<\/sup><\/p>\n<p>Since, 855<sup>o<\/sup>\u00a0lies in the II quadrant in which the cosine function is negative.<\/p>\n<p>cos 855<sup>o<\/sup>\u00a0= cos (90\u00d79 + 45)<sup>o<\/sup><\/p>\n<p>= \u2013 sin 45<sup>o<\/sup><\/p>\n<p>= \u2013 1\/\u221a2<\/p>\n<p><strong>(xii)<\/strong>\u00a0sin 41\u03c0\/4<\/p>\n<p>sin 41\u03c0\/4 = sin 1845<sup>o<\/sup><\/p>\n<p>= sin (90\u00d720 + 45)<sup>o<\/sup><\/p>\n<p>Since, 1845<sup>o<\/sup>\u00a0lies in the I quadrant in which the sine function is positive.<\/p>\n<p>sin 1845<sup>o<\/sup>\u00a0= sin (90\u00d720 + 45)<sup>o<\/sup><\/p>\n<p>= sin 45<sup>o<\/sup><\/p>\n<p>= 1\/\u221a2<\/p>\n<p><strong>(xiii)<\/strong>\u00a0cos 39\u03c0\/4<\/p>\n<p>cos 39\u03c0\/4 = cos 1755<sup>o<\/sup><\/p>\n<p>= cos (90\u00d719 + 45)<sup>o<\/sup><\/p>\n<p>Since, 1755<sup>o<\/sup>\u00a0lies in the IV quadrant in which the cosine function is positive.<\/p>\n<p>cos 1755<sup>o<\/sup>\u00a0= cos (90\u00d719 + 45)<sup>o<\/sup><\/p>\n<p>= sin 45<sup>o<\/sup><\/p>\n<p>= 1\/\u221a2<\/p>\n<p><strong>(xiv)<\/strong>\u00a0sin 151\u03c0\/6<\/p>\n<p>sin 151\u03c0\/6 = sin 4530<sup>o<\/sup><\/p>\n<p>= sin (90\u00d750 + 30)<sup>o<\/sup><\/p>\n<p>Since, 4530<sup>o<\/sup>\u00a0lies in the III quadrant in which the sine function is negative.<\/p>\n<p>sin 4530<sup>o<\/sup>\u00a0= sin (90\u00d750 + 30)<sup>o<\/sup><\/p>\n<p>= \u2013 sin 30<sup>o<\/sup><\/p>\n<p>= -1\/2<\/p>\n<h3><span class=\"ez-toc-section\" id=\"2-prove-that-i-tan-225o-cot-405o-tan-765o-cot-675o-0\"><\/span>2. prove that:<br \/>(i) tan 225<sup>o<\/sup>\u00a0cot 405<sup>o<\/sup>\u00a0+ tan 765<sup>o<\/sup>\u00a0cot 675<sup>o<\/sup>\u00a0= 0<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"ii-sin-8%cf%803-cos-23%cf%806-cos-13%cf%803-sin-35%cf%806-12\"><\/span>(ii) sin 8\u03c0\/3 cos 23\u03c0\/6 + cos 13\u03c0\/3 sin 35\u03c0\/6 = 1\/2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iii-cos-24o-cos-55o-cos-125o-cos-204o-cos-300o-12\"><\/span>(iii) cos 24<sup>o<\/sup>\u00a0+ cos 55<sup>o<\/sup>\u00a0+ cos 125<sup>o<\/sup>\u00a0+ cos 204<sup>o<\/sup>\u00a0+ cos 300<sup>o<\/sup>\u00a0= 1\/2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"iv-tan-125o-cot-405o-%e2%80%93-tan-765o-cot-675o-0\"><\/span>(iv) tan (-125<sup>o<\/sup>) cot (-405<sup>o<\/sup>) \u2013 tan (-765<sup>o<\/sup>) cot (675<sup>o<\/sup>) = 0<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"v-cos-570o-sin-510o-sin-330o-cos-390o-0\"><\/span>(v) cos 570<sup>o<\/sup>\u00a0sin 510<sup>o<\/sup>\u00a0+ sin (-330<sup>o<\/sup>) cos (-390<sup>o<\/sup>) = 0<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"vi-tan-11%cf%803-%e2%80%93-2-sin-4%cf%806-%e2%80%93-34-cosec2-%cf%804-4-cos2-17%cf%806-3-%e2%80%93-4%e2%88%9a32\"><\/span>(vi) tan 11\u03c0\/3 \u2013 2 sin 4\u03c0\/6 \u2013 3\/4 cosec<sup>2<\/sup>\u00a0\u03c0\/4 + 4 cos<sup>2<\/sup>\u00a017\u03c0\/6 = (3 \u2013 4\u221a3)\/2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"vii-3-sin-%cf%806-sec-%cf%803-%e2%80%93-4-sin-5%cf%806-cot-%cf%804-1\"><\/span>(vii) 3 sin \u03c0\/6 sec \u03c0\/3 \u2013 4 sin 5\u03c0\/6 cot \u03c0\/4 = 1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>tan 225<sup>o<\/sup>\u00a0cot 405<sup>o<\/sup>\u00a0+ tan 765<sup>o<\/sup>\u00a0cot 675<sup>o<\/sup>\u00a0= 0<\/p>\n<p>Let us consider LHS:<\/p>\n<p>tan 225\u00b0 cot 405\u00b0 + tan 765\u00b0 cot 675\u00b0<\/p>\n<p>tan (90\u00b0 \u00d7 2 + 45\u00b0) cot (90\u00b0 \u00d7 4 + 45\u00b0) + tan (90\u00b0 \u00d7 8 + 45\u00b0) cot (90\u00b0 \u00d7 7 + 45\u00b0)<\/p>\n<p>We know that when n is odd, cot\u00a0\u2192\u00a0tan.<\/p>\n<p>tan 45\u00b0 cot 45\u00b0 + tan 45\u00b0 [-tan 45\u00b0]<\/p>\n<p>tan 45\u00b0 cot 45\u00b0 \u2013 tan 45\u00b0 tan 45\u00b0<\/p>\n<p>1 \u00d7 1 \u2013 1 \u00d7 1<\/p>\n<p>1 \u2013 1<\/p>\n<p>0 = RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(ii)\u00a0<\/strong>sin 8\u03c0\/3 cos 23\u03c0\/6 + cos 13\u03c0\/3 sin 35\u03c0\/6 = 1\/2<\/p>\n<p>Let us consider LHS:<\/p>\n<p>sin 8\u03c0\/3 cos 23\u03c0\/6 + cos 13\u03c0\/3 sin 35\u03c0\/6<\/p>\n<p>sin 480\u00b0 cos 690\u00b0 + cos 780\u00b0 sin 1050\u00b0<\/p>\n<p>sin (90\u00b0 \u00d7 5 + 30\u00b0) cos (90\u00b0 \u00d7 7 + 60\u00b0) + cos (90\u00b0 \u00d7 8 + 60\u00b0) sin (90\u00b0 \u00d7 11 + 60\u00b0)<\/p>\n<p>We know that when n is odd, sin\u00a0\u2192\u00a0cos and cos\u00a0\u2192\u00a0sin.<\/p>\n<p>cos 30\u00b0 sin 60\u00b0 + cos 60\u00b0 [-cos 60\u00b0]<\/p>\n<p>\u221a3\/2 \u00d7 \u221a3\/2 \u2013 1\/2 \u00d7 1\/2<\/p>\n<p>3\/4 \u2013 1\/4<\/p>\n<p>2\/4<\/p>\n<p>1\/2<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(iii)\u00a0<\/strong>cos 24<sup>o<\/sup>\u00a0+ cos 55<sup>o<\/sup>\u00a0+ cos 125<sup>o<\/sup>\u00a0+ cos 204<sup>o<\/sup>\u00a0+ cos 300<sup>o<\/sup>\u00a0= 1\/2<\/p>\n<p>Let us consider LHS:<\/p>\n<p>cos 24<sup>o<\/sup>\u00a0+ cos 55<sup>o<\/sup>\u00a0+ cos 125<sup>o<\/sup>\u00a0+ cos 204<sup>o<\/sup>\u00a0+ cos 300<sup>o<\/sup><\/p>\n<p>cos 24\u00b0 + cos (90\u00b0 \u00d7 1 \u2013 35\u00b0) + cos (90\u00b0 \u00d7 1 + 35\u00b0) + cos (90\u00b0 \u00d7 2 + 24\u00b0) + cos (90\u00b0 \u00d7 3 + 30\u00b0)<\/p>\n<p>We know that when n is odd, cos\u00a0\u2192\u00a0sin.<\/p>\n<p>cos 24\u00b0 + sin 35\u00b0 \u2013 sin 35\u00b0 \u2013 cos 24\u00b0 + sin 30\u00b0<\/p>\n<p>0 + 0 + 1\/2<\/p>\n<p>1\/2<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(iv)\u00a0<\/strong>tan (-125<sup>o<\/sup>) cot (-405<sup>o<\/sup>) \u2013 tan (-765<sup>o<\/sup>) cot (675<sup>o<\/sup>) = 0<\/p>\n<p>Let us consider LHS:<\/p>\n<p>tan (-125<sup>o<\/sup>) cot (-405<sup>o<\/sup>) \u2013 tan (-765<sup>o<\/sup>) cot (675<sup>o<\/sup>)<\/p>\n<p>We know that tan (-x) = -tan (x) and cot (-x) = -cot (x).<\/p>\n<p>[-tan (225\u00b0)] [-cot (405\u00b0)] \u2013 [-tan (765\u00b0)] cot (675\u00b0)<\/p>\n<p>\u00a0<\/p>\n<p>tan (225\u00b0) cot (405\u00b0) + tan (765\u00b0) cot (675\u00b0)<\/p>\n<p>tan (90\u00b0 \u00d7 2 + 45\u00b0) cot (90\u00b0 \u00d7 4 + 45\u00b0) + tan (90\u00b0 \u00d7 8 + 45\u00b0) cot (90\u00b0 \u00d7 7 + 45\u00b0)<\/p>\n<p>tan 45\u00b0 cot 45\u00b0 + tan 45\u00b0 [-tan 45\u00b0]<\/p>\n<p>1 \u00d7 1 + 1 \u00d7 (-1)<\/p>\n<p>1 \u2013 1<\/p>\n<p>0<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(v)\u00a0<\/strong>cos 570<sup>o<\/sup>\u00a0sin 510<sup>o<\/sup>\u00a0+ sin (-330<sup>o<\/sup>) cos (-390<sup>o<\/sup>) = 0<\/p>\n<p>Let us consider LHS:<\/p>\n<p>cos 570<sup>o<\/sup>\u00a0sin 510<sup>o<\/sup>\u00a0+ sin (-330<sup>o<\/sup>) cos (-390<sup>o<\/sup>)<\/p>\n<p>We know that sin (-x) = -sin (x) and cos (-x) = +cos (x).<\/p>\n<p>cos 570<sup>o<\/sup>\u00a0sin 510<sup>o<\/sup>\u00a0+ [-sin (330<sup>o<\/sup>)] cos (390<sup>o<\/sup>)<\/p>\n<p>cos 570<sup>o<\/sup>\u00a0sin 510<sup>o<\/sup>\u00a0\u2013 sin (330<sup>o<\/sup>) cos (390<sup>o<\/sup>)<\/p>\n<p>cos (90\u00b0 \u00d7 6 + 30\u00b0) sin (90\u00b0 \u00d7 5 + 60\u00b0) \u2013 sin (90\u00b0 \u00d7 3 + 60\u00b0) cos (90\u00b0 \u00d7 4 + 30\u00b0)<\/p>\n<p>We know that cos is negative at 90\u00b0 + \u03b8 i.e. in Q<sub>2<\/sub>\u00a0and when n is odd, sin\u00a0\u2192\u00a0cos and cos\u00a0\u2192\u00a0sin.<\/p>\n<p>-cos 30\u00b0 cos 60\u00b0 \u2013 [-cos 60\u00b0] cos 30\u00b0<\/p>\n<p>-cos 30\u00b0 cos 60\u00b0 + cos 60\u00b0 cos 30\u00b0<\/p>\n<p>0<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(vi)\u00a0<\/strong>tan 11\u03c0\/3 \u2013 2 sin 4\u03c0\/6 \u2013 3\/4 cosec<sup>2<\/sup>\u00a0\u03c0\/4 + 4 cos<sup>2<\/sup>\u00a017\u03c0\/6 = (3 \u2013 4\u221a3)\/2<\/p>\n<p>Let us consider LHS:<\/p>\n<p>tan 11\u03c0\/3 \u2013 2 sin 4\u03c0\/6 \u2013 3\/4 cosec<sup>2<\/sup>\u00a0\u03c0\/4 + 4 cos<sup>2<\/sup>\u00a017\u03c0\/6<\/p>\n<p>tan (11 \u00d7 180<sup>o<\/sup>)\/3 \u2013 2 sin (4 \u00d7 180<sup>o<\/sup>)\/6 \u2013 3\/4 cosec<sup>2<\/sup>\u00a0180<sup>o<\/sup>\/4 + 4 cos<sup>2<\/sup>\u00a0(17 \u00d7 180<sup>o<\/sup>)\/6<\/p>\n<p>tan 660<sup>o<\/sup>\u00a0\u2013 2 sin 120<sup>o<\/sup>\u00a0\u2013 3\/4 (cosec 45<sup>o<\/sup>)<sup>2<\/sup>\u00a0+ 4 (cos 510<sup>o<\/sup>)<sup>2<\/sup><\/p>\n<p>tan (90\u00b0 \u00d7 7 + 30\u00b0) \u2013 2 sin (90\u00b0 \u00d7 1 + 30\u00b0) \u2013 3\/4 [cosec 45\u00b0]<sup>2<\/sup>\u00a0+ 4 [cos (90\u00b0 \u00d7 5 + 60\u00b0)]<sup>2<\/sup><\/p>\n<p>We know that tan and cos is negative at 90\u00b0 + \u03b8 i.e. in Q<sub>2<\/sub> and when n is odd, tan \u2192 cot, sin \u2192 cos, and cos \u2192 sin.<\/p>\n<p>[-cot 30\u00b0] \u2013 2 cos 30\u00b0 \u2013 3\/4 [cosec 45\u00b0]<sup>2<\/sup>\u00a0+ [-sin 60\u00b0]<sup>2<\/sup><\/p>\n<p>\u00a0<\/p>\n<p>\u2013 cot 30\u00b0 \u2013 2 cos 30\u00b0 \u2013 3\/4 [cosec 45\u00b0]<sup>2<\/sup>\u00a0+ [sin 60\u00b0]<sup>2<\/sup><\/p>\n<p>-\u221a3 \u2013 2\u221a3\/2 \u2013 3\/4 (\u221a2)<sup>2<\/sup>\u00a0+ 4 (\u221a3\/2)<sup>2<\/sup><\/p>\n<p>-\u221a3 \u2013 \u221a3 \u2013 6\/4 + 12\/4<\/p>\n<p>(3 \u2013 4\u221a3)\/2<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(vii)\u00a0<\/strong>3 sin \u03c0\/6 sec \u03c0\/3 \u2013 4 sin 5\u03c0\/6 cot \u03c0\/4 = 1<\/p>\n<p>Let us consider LHS:<\/p>\n<p>3 sin \u03c0\/6 sec \u03c0\/3 \u2013 4 sin 5\u03c0\/6 cot \u03c0\/4<\/p>\n<p>3 sin 180<sup>o<\/sup>\/6 sec 180<sup>o<\/sup>\/3 \u2013 4 sin 5(180<sup>o<\/sup>)\/6 cot 180<sup>o<\/sup>\/4<\/p>\n<p>3 sin 30\u00b0 sec 60\u00b0 \u2013 4 sin 150\u00b0 cot 45\u00b0<\/p>\n<p>3 sin 30\u00b0 sec 60\u00b0 \u2013 4 sin (90\u00b0 \u00d7 1 + 60\u00b0) cot 45\u00b0<\/p>\n<p>We know that when n is odd, sin\u00a0\u2192\u00a0cos.<\/p>\n<p>3 sin 30\u00b0 sec 60\u00b0 \u2013 4 cos 60\u00b0 cot 45\u00b0<\/p>\n<p>3 (1\/2) (2) \u2013 4 (1\/2) (1)<\/p>\n<p>3 \u2013 2<\/p>\n<p>1<\/p>\n<p>= RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"3-prove-that\"><\/span>3. Prove that:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"i\"><\/span>(i)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><img class=\"alignnone size-full wp-image-118117\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-5-5.gif\" alt=\"\" width=\"338\" height=\"44\"><\/p>\n<h3><span class=\"ez-toc-section\" id=\"ii\"><\/span><strong>(<\/strong>ii<strong>)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>\u00a0<\/p>\n<p><img class=\"alignnone size-full wp-image-118123\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-5-6.gif\" alt=\"\" width=\"535\" height=\"43\"><\/p>\n<h3><span class=\"ez-toc-section\" id=\"iii\"><\/span>(iii)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><img class=\"alignnone size-full wp-image-118126\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-5-7.gif\" alt=\"\" width=\"397\" height=\"49\"><\/p>\n<h3><span class=\"ez-toc-section\" id=\"iv\"><\/span><strong>(<\/strong>iv<strong>)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><img class=\"alignnone size-full wp-image-118131\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-5-8.gif\" alt=\"\" width=\"470\" height=\"34\"><\/p>\n<h3><span class=\"ez-toc-section\" id=\"v\"><\/span><strong>(<\/strong>v<strong>)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><img class=\"alignnone size-full wp-image-118135\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-5-9.gif\" alt=\"\" width=\"310\" height=\"46\"><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-118138\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-528.png\" alt=\"\" width=\"349\" height=\"258\"><\/p>\n<p>1 = RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(ii)<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-118139\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-529.png\" alt=\"\" width=\"574\" height=\"354\"><\/p>\n<p>1 + 1<\/p>\n<p>2 = RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(iii)<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-118143\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-531.png\" alt=\"\" width=\"537\" height=\"345\"><\/p>\n<p>1 = RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(iv)<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-118144\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-532.png\" alt=\"\" width=\"481\" height=\"105\"><\/p>\n<p>{1 + cot x \u2013 (-cosec x)} {1 + cot x + (-cosec x)}<\/p>\n<p>{1 + cot x + cosec x} {1 + cot x \u2013 cosec x}<\/p>\n<p>{(1 + cot x) + (cosec x)} {(1 + cot x) \u2013 (cosec x)}<\/p>\n<p>By using the formula, (a + b) (a \u2013 b) = a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup><\/p>\n<p>(1 + cot x)<sup>2<\/sup>\u00a0\u2013 (cosec x)<sup>2<\/sup><\/p>\n<p>1 + cot<sup>2\u00a0<\/sup>x + 2 cot x \u2013 cosec<sup>2\u00a0<\/sup>x<\/p>\n<p>We know that 1 + cot<sup>2\u00a0<\/sup>x = cosec<sup>2\u00a0<\/sup>x<\/p>\n<p>cosec<sup>2\u00a0<\/sup>x + 2 cot x \u2013 cosec<sup>2\u00a0<\/sup>x<\/p>\n<p>2 cot x = RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>(v)<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-118145\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/rd-533.png\" alt=\"\" width=\"445\" height=\"460\"><\/p>\n<p>1 = RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>4. Prove that: sin<sup>2<\/sup>\u00a0\u03c0\/18 + sin<sup>2<\/sup>\u00a0\u03c0\/9 + sin<sup>2<\/sup>\u00a07\u03c0\/18 + sin<sup>2<\/sup>\u00a04\u03c0\/9 = 2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>sin<sup>2<\/sup>\u00a0\u03c0\/18 + sin<sup>2<\/sup>\u00a0\u03c0\/9 + sin<sup>2<\/sup>\u00a07\u03c0\/18 + sin<sup>2<\/sup>\u00a04\u03c0\/9<\/p>\n<p>sin<sup>2<\/sup>\u00a0\u03c0\/18 + sin<sup>2<\/sup>\u00a02\u03c0\/18 + sin<sup>2<\/sup>\u00a07\u03c0\/18 + sin<sup>2<\/sup>\u00a08\u03c0\/18<\/p>\n<p>sin<sup>2<\/sup>\u00a0\u03c0\/18 + sin<sup>2<\/sup>\u00a02\u03c0\/18 + sin<sup>2<\/sup>\u00a0(\u03c0\/2 \u2013 2\u03c0\/18) + sin<sup>2<\/sup>\u00a0(\u03c0\/2 \u2013 \u03c0\/18)<\/p>\n<p>We know that when n is odd, sin\u00a0\u2192\u00a0cos.<\/p>\n<p>sin<sup>2<\/sup>\u00a0\u03c0\/18 + sin<sup>2<\/sup>\u00a02\u03c0\/18 + cos<sup>2<\/sup>\u00a02\u03c0\/18 + cos<sup>2<\/sup>\u00a02\u03c0\/18<\/p>\n<p>when rearranged,<\/p>\n<p>sin<sup>2<\/sup>\u00a0\u03c0\/18 + cos<sup>2<\/sup>\u00a02\u03c0\/18 + sin<sup>2<\/sup>\u00a0\u03c0\/18 + cos<sup>2<\/sup>\u00a02\u03c0\/18<\/p>\n<p>We know that sin<sup>2<\/sup>\u00a0+ cos<sup>2<\/sup>x = 1.<\/p>\n<p>So,<\/p>\n<p>1 + 1<\/p>\n<p>2 = RHS<\/p>\n<p>\u2234 LHS = RHS<\/p>\n<p>Hence proved.<\/p>\n<p><span style=\"font-size: 30px; font-weight: bold; background-color: initial;\">Importa<\/span><span style=\"font-size: 30px; font-weight: bold; background-color: initial;\">nt Topics Class 11 Maths Chapter 23<\/span><\/p>\n<p>RD Sharma Solutions Class 11 Maths Chapter 5, undoubtedly is one of the chapters that students need to take seriously from the very beginning of the course curriculum. So, to make it better, have a quick look at the important topics that are there, that need to be discussed several times before appearing for the exam. Besides helping you with scoring well, these <span style=\"font-weight: 400;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\">RD Sharma Solutions<\/a><\/span> will let you enhance your exam performance.<\/p>\n<ul>\n<li>Trigonometric functions of a real number.<\/li>\n<li>Values of trigonometric functions.<\/li>\n<li>Trigonometric identities.<\/li>\n<li>Fundamental trigonometric identities.<\/li>\n<li>Signs of trigonometric functions.<\/li>\n<li>Variations in values of trigonometric functions in different quadrants.<\/li>\n<li>Values of trigonometric functions at allied angles.<\/li>\n<li>Periodic functions.<\/li>\n<li>Even and odd functions.<\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"access-other-important-chapters-of-rd-sharma-solutions-class-11-maths\"><\/span>Access Other Important Chapters of RD Sharma Solutions Class 11 Maths\u00a0<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 1 &#8211; Sets<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 2 &#8211; Relations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 3 &#8211; Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 4 &#8211; Measurement of Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 6 &#8211; Graphs of Trigonometric Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 7 &#8211; Trigonometric Ratios of Compound Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 8 &#8211; Transformation Formulae<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 9 &#8211; Trigonometric Ratios of Multiple and Sub-Multiple Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 10 &#8211; Sine and Cosine Formulae and Their Applications<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 11 &#8211; Trigonometric Equations<\/a><\/li>\n<\/ul>\n<p>Hence paying focus in this chapter is what you need to do right now. Have your copy of RD Sharma Solutions <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener noreferrer\">CBSE<\/a> Class 11 Maths Chapter 5 PDF and begin your exam preparation.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-11-maths-chapter-5\"><\/span>FAQs on RD Sharma Solutions Class 11 Maths Chapter 5<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-1629793658792\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-rd-sharma-solutions-class-11-maths-chapter-5-pdf\"><\/span>From where can I download the RD Sharma Solutions Class 11 Maths Chapter 5 PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link in the above blog. <\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629793826279\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-5\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 5?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download the PDF for free. <\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629794226317\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"are-the-solutions-of-rd-sharma-solutions-class-11-maths-chapter-5-reliable\"><\/span>Are the solutions of RD Sharma Solutions Class 11 Maths Chapter 5 reliable?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>The solutions are prepared by subject matter experts, hence reliable. <\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 5-Trigonometric Functions: RD Sharma Solutions Class 11 Maths Chapter 5 is here for you to prepare the 5th chapter (Trigonometric Functions) from your class 11 Mathematics syllabus. Chapter 5 from class 11 talks about Trigonometric Functions which is one of the Important chapters which needs clarity. And with &#8230; <a title=\"RD Sharma Solutions Class 11 Maths Chapter 5 &#8211; Trigonometric Functions (Updated For 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-5\/\" aria-label=\"More on RD Sharma Solutions Class 11 Maths Chapter 5 &#8211; Trigonometric Functions (Updated For 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":119052,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,73411,73410],"tags":[3428,73334],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62362"}],"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\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=62362"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62362\/revisions"}],"predecessor-version":[{"id":504789,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62362\/revisions\/504789"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/119052"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=62362"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=62362"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=62362"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}