{"id":544672,"date":"2021-10-02T08:26:59","date_gmt":"2021-10-02T08:26:59","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=544672"},"modified":"2021-10-04T06:10:39","modified_gmt":"2021-10-04T06:10:39","slug":"rd-sharma-solutions-for-class-10-maths-chapter-6-trigonometric-identities","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-6-trigonometric-identities\/","title":{"rendered":"RD Sharma Solutions for Class 10 Maths Chapter 6\u2013Trigonometric Identities"},"content":{"rendered":"\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Class 10: Maths Chapter 6 solutions. Complete Class 10 Maths Chapter 6 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-rd-sharma-solutions-for-class-10-maths-chapter-6-trigonometric-identities\">RD Sharma Solutions for Class 10 Maths Chapter 6\u2013Trigonometric Identities<\/h2>\n\n\n\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Class 10: Maths Chapter 6 solutions. Complete Class 10 Maths Chapter 6 Notes.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 6.1 Page No: 6.43<\/h4>\n\n\n\n<p><strong>Prove the following trigonometric identities:<\/strong><\/p>\n\n\n\n<p><strong>1. (1 \u2013 cos<sup>2<\/sup>&nbsp;A) cosec<sup>2<\/sup>&nbsp;A = 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking the L.H.S,<\/p>\n\n\n\n<p>(1 \u2013 cos<sup>2<\/sup>&nbsp;A) cosec<sup>2<\/sup>&nbsp;A<\/p>\n\n\n\n<p>= (sin<sup>2<\/sup>&nbsp;A) cosec<sup>2<\/sup>&nbsp;A [\u2235 sin<sup>2<\/sup>&nbsp;A + cos<sup>2&nbsp;<\/sup>A = 1 \u21d21 \u2013 sin<sup>2<\/sup>&nbsp;A = cos<sup>2<\/sup>&nbsp;A]<\/p>\n\n\n\n<p>= 1<sup>2<\/sup><\/p>\n\n\n\n<p>= 1 = R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>2. (1 + cot<sup>2<\/sup>&nbsp;A) sin<sup>2<\/sup>&nbsp;A = 1&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>Solution:&nbsp;<\/strong><\/p>\n\n\n\n<p>By using the identity,<\/p>\n\n\n\n<p>cosec<sup>2&nbsp;<\/sup>A \u2013 cot<sup>2<\/sup>&nbsp;A = 1 \u21d2 cosec<sup>2&nbsp;<\/sup>A = cot<sup>2<\/sup>&nbsp;A + 1<\/p>\n\n\n\n<p>Taking,<\/p>\n\n\n\n<p>L.H.S = (1 + cot<sup>2<\/sup>&nbsp;A) sin<sup>2<\/sup>&nbsp;A<\/p>\n\n\n\n<p>= cosec<sup>2<\/sup>&nbsp;A sin<sup>2<\/sup>&nbsp;A<\/p>\n\n\n\n<p>= (cosec A sin A)<sup>2<\/sup><\/p>\n\n\n\n<p>= ((1\/sin A) \u00d7 sin A)<sup>2<\/sup><\/p>\n\n\n\n<p>= (1)<sup>2<\/sup><\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>3. tan<sup>2&nbsp;<\/sup>\u03b8 cos<sup>2&nbsp;<\/sup>\u03b8 =&nbsp;1 \u2212 cos<sup>2&nbsp;<\/sup>\u03b8&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>Solution:&nbsp;<\/strong><\/p>\n\n\n\n<p>We know that,<\/p>\n\n\n\n<p>sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>Taking,<\/p>\n\n\n\n<p>L.H.S =&nbsp;tan<sup>2&nbsp;<\/sup>\u03b8 cos<sup>2&nbsp;<\/sup>\u03b8<\/p>\n\n\n\n<p>= (tan \u03b8 \u00d7 cos \u03b8)<sup>2<\/sup><\/p>\n\n\n\n<p>= (sin \u03b8)<sup>2<\/sup><\/p>\n\n\n\n<p>= sin<sup>2&nbsp;<\/sup>\u03b8<\/p>\n\n\n\n<p>= 1 \u2013 cos<sup>2&nbsp;<\/sup>\u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>4. cosec \u03b8 \u221a(1 \u2013 cos<sup>2<\/sup>&nbsp;\u03b8) = 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Using identity,<\/p>\n\n\n\n<p>sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1&nbsp; \u21d2 sin<sup>2&nbsp;<\/sup>\u03b8 = 1 \u2013 cos<sup>2&nbsp;<\/sup>\u03b8<\/p>\n\n\n\n<p>Taking L.H.S,<\/p>\n\n\n\n<p>L.H.S = cosec \u03b8 \u221a(1 \u2013 cos<sup>2<\/sup>&nbsp;\u03b8)<\/p>\n\n\n\n<p>= cosec \u03b8 \u221a( sin<sup>2&nbsp;<\/sup>\u03b8)<\/p>\n\n\n\n<p>= cosec \u03b8 x sin<sup>&nbsp;<\/sup>\u03b8<\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>5. (sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 1)(cosec<sup>2&nbsp;<\/sup>\u03b8 \u2212 1) = 1&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Using identities,<\/p>\n\n\n\n<p>(sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 tan<sup>2&nbsp;<\/sup>\u03b8) = 1&nbsp;and (cosec<sup>2&nbsp;<\/sup>\u03b8 \u2212 cot<sup>2&nbsp;<\/sup>\u03b8) = 1<\/p>\n\n\n\n<p>We have,<\/p>\n\n\n\n<p>L.H.S =&nbsp;(sec<sup>2&nbsp;<\/sup>\u03b8 \u2013 1)(cosec<sup>2<\/sup>\u03b8 \u2013 1)<\/p>\n\n\n\n<p>= tan<sup>2<\/sup>\u03b8 \u00d7 cot<sup>2<\/sup>\u03b8<\/p>\n\n\n\n<p>= (tan \u03b8 \u00d7 cot \u03b8)<sup>2<\/sup><\/p>\n\n\n\n<p>= (tan \u03b8 \u00d7 1\/tan \u03b8)<sup>2<\/sup><\/p>\n\n\n\n<p>= 1<sup>2<\/sup><\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>6. tan \u03b8 + 1\/ tan \u03b8 = sec \u03b8 cosec \u03b8<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We have,<\/p>\n\n\n\n<p>L.H.S = tan \u03b8 + 1\/ tan \u03b8<\/p>\n\n\n\n<p>= (tan<sup>2<\/sup>&nbsp;\u03b8 + 1)\/ tan \u03b8<\/p>\n\n\n\n<p>= sec<sup>2<\/sup>&nbsp;\u03b8 \/ tan \u03b8 [\u2235 sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 tan<sup>2&nbsp;<\/sup>\u03b8 = 1]<\/p>\n\n\n\n<p>= (1\/cos<sup>2<\/sup>&nbsp;\u03b8) x 1\/ (sin \u03b8\/cos \u03b8) [\u2235 tan \u03b8 = sin \u03b8 \/ cos \u03b8]<\/p>\n\n\n\n<p>= cos \u03b8\/ (sin \u03b8 x cos<sup>2<\/sup>&nbsp;\u03b8)<\/p>\n\n\n\n<p>= 1\/ cos \u03b8 x 1\/ sin \u03b8<\/p>\n\n\n\n<p>= sec \u03b8 x cosec \u03b8<\/p>\n\n\n\n<p>= sec \u03b8 cosec \u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>7. cos \u03b8\/ (1 \u2013 sin \u03b8) = (1 + sin \u03b8)\/ cos \u03b8<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We know that,<\/p>\n\n\n\n<p>sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>So, by multiplying both the numerator and the denominator by (1+ sin \u03b8), we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"221\" height=\"296\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6.png\" alt=\"\" class=\"wp-image-544676\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6.png 221w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-150x200.png 150w\" sizes=\"auto, (max-width: 221px) 100vw, 221px\" \/><\/figure>\n\n\n\n<p>L.H.S =<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p>8.&nbsp;<strong>cos \u03b8\/ (1 + sin \u03b8) = (1 \u2013 sin \u03b8)\/ cos \u03b8<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We know that,<\/p>\n\n\n\n<p>sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>So, by multiplying both the numerator and the denominator by (1- sin \u03b8), we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"200\" height=\"375\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-1.png\" alt=\"\" class=\"wp-image-544677\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-1.png 200w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-1-160x300.png 160w\" sizes=\"auto, (max-width: 200px) 100vw, 200px\" \/><\/figure>\n\n\n\n<p>L.H.S =<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>9. cos<sup>2&nbsp;<\/sup>\u03b8 + 1\/(1 + cot<sup>2&nbsp;<\/sup>\u03b8) = 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We already know that,<\/p>\n\n\n\n<p>cosec<sup>2&nbsp;<\/sup>\u03b8 \u2212 cot<sup>2&nbsp;<\/sup>\u03b8 = 1 and sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>Taking L.H.S,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"228\" height=\"176\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-2.png\" alt=\"\" class=\"wp-image-544678\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= cos<sup>2<\/sup>&nbsp;A + sin<sup>2<\/sup>&nbsp;A<\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>10. sin<sup>2&nbsp;<\/sup>A + 1\/(1 + tan&nbsp;<sup>2&nbsp;<\/sup>A) = 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We already know that,<\/p>\n\n\n\n<p>sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 tan<sup>2&nbsp;<\/sup>\u03b8 = 1&nbsp;and sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>Taking L.H.S,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"224\" height=\"170\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-3.png\" alt=\"\" class=\"wp-image-544679\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= sin<sup>2<\/sup>&nbsp;A + cos<sup>2<\/sup>&nbsp;A<\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>11.<\/strong><br><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"234\" height=\"71\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-4.png\" alt=\"\" class=\"wp-image-544680\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We know that, sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>Taking the L.H.S,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"264\" height=\"348\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-5.png\" alt=\"\" class=\"wp-image-544681\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-5.png 264w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-5-228x300.png 228w\" sizes=\"auto, (max-width: 264px) 100vw, 264px\" \/><\/figure>\n\n\n\n<p>= cosec \u03b8 \u2013 cot&nbsp;\u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>12. 1 \u2013 cos \u03b8\/ sin \u03b8 = sin \u03b8\/ 1 + cos \u03b8<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We know that,<\/p>\n\n\n\n<p>sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>So, by multiplying both the numerator and the denominator by (1+ cos \u03b8), we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"225\" height=\"190\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-6.png\" alt=\"\" class=\"wp-image-544682\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>13.<\/strong>&nbsp;<strong>sin \u03b8\/ (1 \u2013 cos \u03b8) = cosec \u03b8 + cot \u03b8<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"270\" height=\"441\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-7.png\" alt=\"\" class=\"wp-image-544683\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-7.png 270w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-7-184x300.png 184w\" sizes=\"auto, (max-width: 270px) 100vw, 270px\" \/><\/figure>\n\n\n\n<p>= cosec \u03b8 + cot \u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>14. (1 \u2013 sin \u03b8) \/ (1 + sin \u03b8) = (sec \u03b8 \u2013 tan \u03b8)<sup>2<\/sup><\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking the L.H.S,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"251\" height=\"390\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-8.png\" alt=\"\" class=\"wp-image-544684\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-8.png 251w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-8-193x300.png 193w\" sizes=\"auto, (max-width: 251px) 100vw, 251px\" \/><\/figure>\n\n\n\n<p>= (sec \u03b8 \u2013 tan \u03b8)<sup>2<\/sup><\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>15.&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"234\" height=\"265\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-10.png\" alt=\"\" class=\"wp-image-544685\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= cot \u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>16. tan<sup>2&nbsp;<\/sup>\u03b8 \u2212 sin<sup>2&nbsp;<\/sup>\u03b8&nbsp;=&nbsp;tan<sup>2&nbsp;<\/sup>\u03b8 sin<sup>2&nbsp;<\/sup>\u03b8&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S,<\/p>\n\n\n\n<p>L.H.S = tan<sup>2&nbsp;<\/sup>\u03b8 \u2212 sin<sup>2&nbsp;<\/sup>\u03b8<strong>&nbsp;<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"186\" height=\"301\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-11.png\" alt=\"\" class=\"wp-image-544686\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= tan<sup>2&nbsp;<\/sup>\u03b8 sin<sup>2&nbsp;<\/sup>\u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>17. (cosec&nbsp;\u03b8&nbsp;+ sin&nbsp;\u03b8)(cosec&nbsp;\u03b8&nbsp;\u2013 sin&nbsp;\u03b8) =&nbsp;cot<sup>2<\/sup>\u03b8 + cos<sup>2<\/sup>\u03b8&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S = (cosec&nbsp;\u03b8&nbsp;+ sin&nbsp;\u03b8)(cosec&nbsp;\u03b8&nbsp;\u2013 sin&nbsp;\u03b8)<\/p>\n\n\n\n<p>On multiplying we get,<\/p>\n\n\n\n<p>= cosec<sup>2<\/sup>&nbsp;\u03b8&nbsp;\u2013 sin<sup>2<\/sup>&nbsp;\u03b8<\/p>\n\n\n\n<p>=&nbsp;(1 + cot<sup>2<\/sup>&nbsp;\u03b8)&nbsp;\u2013 (1 \u2013 cos<sup>2<\/sup>&nbsp;\u03b8) [Using cosec<sup>2&nbsp;<\/sup>\u03b8 \u2212 cot<sup>2&nbsp;<\/sup>\u03b8 = 1&nbsp;and sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1]<\/p>\n\n\n\n<p>=&nbsp;1 + cot<sup>2<\/sup>&nbsp;\u03b8&nbsp;\u2013 1 + cos<sup>2<\/sup>&nbsp;\u03b8<\/p>\n\n\n\n<p>= cot<sup>2<\/sup>&nbsp;\u03b8&nbsp;+ cos<sup>2<\/sup>&nbsp;\u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>18. (sec \u03b8&nbsp;+ cos&nbsp;\u03b8) (sec&nbsp;\u03b8&nbsp;\u2013 cos&nbsp;\u03b8) =&nbsp;tan<sup>2&nbsp;<\/sup>\u03b8 + sin<sup>2&nbsp;<\/sup>\u03b8&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S = (sec&nbsp;\u03b8&nbsp;+ cos&nbsp;\u03b8)(sec&nbsp;\u03b8&nbsp;\u2013 cos&nbsp;\u03b8)<\/p>\n\n\n\n<p>On multiplying we get,<\/p>\n\n\n\n<p>= sec<sup>2<\/sup>&nbsp;\u03b8&nbsp;\u2013 sin<sup>2<\/sup>&nbsp;\u03b8<\/p>\n\n\n\n<p>= (1 + tan<sup>2<\/sup>&nbsp;\u03b8)&nbsp;\u2013 (1 \u2013 sin<sup>2<\/sup>&nbsp;\u03b8) [Using sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 tan<sup>2&nbsp;<\/sup>\u03b8 = 1&nbsp;and sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1]<\/p>\n\n\n\n<p>= 1 + tan<sup>2<\/sup>&nbsp;\u03b8&nbsp;\u2013 1 + sin<sup>2<\/sup>&nbsp;\u03b8<\/p>\n\n\n\n<p>=&nbsp;tan&nbsp;<sup>2&nbsp;<\/sup>\u03b8&nbsp;+ sin&nbsp;<sup>2<\/sup>&nbsp;\u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>19. sec A(1- sin A) (sec A + tan A) = 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S = sec A(1 \u2013 sin A)(sec A + tan A)<\/p>\n\n\n\n<p>Substituting sec A = 1\/cos A and tan A =sin A\/cos A in the above we have,<\/p>\n\n\n\n<p>L.H.S = 1\/cos A (1 \u2013 sin A)(1\/cos A + sin A\/cos A)<\/p>\n\n\n\n<p>= 1 \u2013 sin<sup>2<\/sup>&nbsp;A \/ cos<sup>2<\/sup>&nbsp;A [After taking L.C.M]<\/p>\n\n\n\n<p>= cos<sup>2<\/sup>&nbsp;A \/ cos<sup>2<\/sup>&nbsp;A [\u2235 1 \u2013 sin<sup>2<\/sup>&nbsp;A = cos<sup>2<\/sup>&nbsp;A]<\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>20. (cosec A \u2013 sin A)(sec A \u2013 cos A)(tan A + cot A) = 1&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S = (cosec A \u2013 sin A)(sec A \u2013 cos A)(tan A + cot A)<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"514\" height=\"239\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-12.png\" alt=\"\" class=\"wp-image-544687\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-12.png 514w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-12-300x139.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-12-400x186.png 400w\" sizes=\"auto, (max-width: 514px) 100vw, 514px\" \/><\/figure>\n\n\n\n<p>= (cos<sup>2<\/sup>&nbsp;A\/ sin A) (sin<sup>2<\/sup>&nbsp;A\/ cos A) (1\/ sin A cos A) [\u2235 sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1]<\/p>\n\n\n\n<p>= (sin A&nbsp; cos A)&nbsp; (1\/ cos A sin A)<\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>21. (1 +&nbsp;tan<sup>2&nbsp;<\/sup>\u03b8)(1 \u2013 sin&nbsp;\u03b8)(1 + sin&nbsp;\u03b8) = 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S = (1 +&nbsp;tan<sup>2<\/sup>\u03b8)(1 \u2013 sin \u03b8)(1 + sin \u03b8)<\/p>\n\n\n\n<p>And, we know sin<sup>2<\/sup>&nbsp;\u03b8&nbsp;+ cos<sup>2<\/sup>&nbsp;\u03b8&nbsp;= 1 and sec<sup>2<\/sup>&nbsp;\u03b8&nbsp;\u2013 tan<sup>2<\/sup>&nbsp;\u03b8&nbsp;= 1<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>L.H.S = (1 +&nbsp;tan<sup>2<\/sup>&nbsp;\u03b8)(1 \u2013 sin&nbsp;\u03b8)(1 + sin&nbsp;\u03b8)<\/p>\n\n\n\n<p>= (1 +&nbsp;tan<sup>2&nbsp;<\/sup>\u03b8){(1 \u2013 sin&nbsp;\u03b8)(1 + sin&nbsp;\u03b8)}<\/p>\n\n\n\n<p>= (1 +&nbsp;tan<sup>2&nbsp;<\/sup>\u03b8)(1 \u2013&nbsp;sin<sup>2&nbsp;<\/sup>\u03b8)<\/p>\n\n\n\n<p>= sec<sup>2&nbsp;<\/sup>\u03b8 (cos<sup>2<\/sup>&nbsp;\u03b8)<\/p>\n\n\n\n<p>= (1\/ cos<sup>2<\/sup>&nbsp;\u03b8) x cos<sup>2<\/sup>&nbsp;\u03b8<\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>22. sin<sup>2&nbsp;<\/sup>A cot<sup>2&nbsp;<\/sup>A + cos<sup>2&nbsp;<\/sup>A tan<sup>2&nbsp;<\/sup>A&nbsp;= 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We know that,<\/p>\n\n\n\n<p>cot<sup>2&nbsp;<\/sup>A = cos<sup>2&nbsp;<\/sup>A\/ sin<sup>2<\/sup>&nbsp;A and tan<sup>2<\/sup>&nbsp;A = sin<sup>2<\/sup>&nbsp;A\/cos<sup>2<\/sup>&nbsp;A<\/p>\n\n\n\n<p>Substituting the above in L.H.S, we get<\/p>\n\n\n\n<p>L.H.S = sin<sup>2&nbsp;<\/sup>A cot<sup>2&nbsp;<\/sup>A + cos<sup>2&nbsp;<\/sup>A tan<sup>2&nbsp;<\/sup>A<\/p>\n\n\n\n<p>= {sin<sup>2&nbsp;<\/sup>A (cos<sup>2&nbsp;<\/sup>A\/ sin<sup>2<\/sup>&nbsp;A)} + {cos<sup>2&nbsp;<\/sup>A (sin<sup>2<\/sup>&nbsp;A\/cos<sup>2<\/sup>&nbsp;A)}<\/p>\n\n\n\n<p>= cos<sup>2&nbsp;<\/sup>A + sin<sup>2<\/sup>&nbsp;A<\/p>\n\n\n\n<p>= 1 [\u2235 sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1]<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"262\" height=\"123\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-13.png\" alt=\"\" class=\"wp-image-544688\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p><strong>23.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Taking the L.H.S and using sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1, we have<\/p>\n\n\n\n<p>L.H.S = cot \u03b8 \u2013 tan \u03b8<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"189\" height=\"307\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-14.png\" alt=\"\" class=\"wp-image-544689\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-14.png 189w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-14-185x300.png 185w\" sizes=\"auto, (max-width: 189px) 100vw, 189px\" \/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p>(ii) Taking the L.H.S and using sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1, we have<\/p>\n\n\n\n<p>L.H.S = tan \u03b8 \u2013 cot \u03b8<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"183\" height=\"301\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-15.png\" alt=\"\" class=\"wp-image-544690\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<p>\u2013 Hence Proved<\/p>\n\n\n\n<p><strong>24. (cos<sup>2<\/sup>&nbsp;\u03b8\/ sin \u03b8) \u2013 cosec \u03b8 + sin \u03b8 = 0<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S and using sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1, we have<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"248\" height=\"327\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-16.png\" alt=\"\" class=\"wp-image-544691\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-16.png 248w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-16-228x300.png 228w\" sizes=\"auto, (max-width: 248px) 100vw, 248px\" \/><\/figure>\n\n\n\n<p>= \u2013 sin \u03b8 + sin \u03b8<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>25.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"433\" height=\"299\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-18.png\" alt=\"\" class=\"wp-image-544692\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-18.png 433w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-18-300x207.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-18-400x276.png 400w\" sizes=\"auto, (max-width: 433px) 100vw, 433px\" \/><\/figure>\n\n\n\n<p>= 2 sec<sup>2<\/sup>&nbsp;A<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>26.&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking the LHS and using sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1, we have<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"252\" height=\"249\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-20.png\" alt=\"\" class=\"wp-image-544693\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= 2\/ cos \u03b8<\/p>\n\n\n\n<p>= 2 sec \u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"296\" height=\"59\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-21.png\" alt=\"\" class=\"wp-image-544694\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p><strong>27.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking the LHS and using sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1, we have<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"409\" height=\"367\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-22.png\" alt=\"\" class=\"wp-image-544695\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-22.png 409w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-22-300x269.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-22-400x359.png 400w\" sizes=\"auto, (max-width: 409px) 100vw, 409px\" \/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>28.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"266\" height=\"58\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-23.png\" alt=\"\" class=\"wp-image-544696\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"91\" height=\"71\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-24.png\" alt=\"\" class=\"wp-image-544697\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>Using sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 tan<sup>2&nbsp;<\/sup>\u03b8 = 1&nbsp;and cosec<sup>2&nbsp;<\/sup>\u03b8 \u2212 cot<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"210\" height=\"108\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-25.png\" alt=\"\" class=\"wp-image-544698\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"173\" height=\"54\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-28.png\" alt=\"\" class=\"wp-image-544699\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p><strong>29.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S and using sin<sup>2&nbsp;<\/sup>\u03b8 + cos<sup>2&nbsp;<\/sup>\u03b8 = 1, we have<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"187\" height=\"44\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-29.png\" alt=\"\" class=\"wp-image-544700\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"237\" height=\"397\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-30.png\" alt=\"\" class=\"wp-image-544701\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-30.png 237w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-30-179x300.png 179w\" sizes=\"auto, (max-width: 237px) 100vw, 237px\" \/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"321\" height=\"57\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-31.png\" alt=\"\" class=\"wp-image-544702\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-31.png 321w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-31-300x53.png 300w\" sizes=\"auto, (max-width: 321px) 100vw, 321px\" \/><\/figure>\n\n\n\n<p><strong>30.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking LHS, we have<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"682\" height=\"450\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-32.png\" alt=\"\" class=\"wp-image-544703\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-32.png 682w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-32-300x198.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-32-400x264.png 400w\" sizes=\"auto, (max-width: 682px) 100vw, 682px\" \/><\/figure>\n\n\n\n<p>= 1 + tan \u03b8 + cot \u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>31. sec<sup>6&nbsp;<\/sup>\u03b8 = tan<sup>6&nbsp;<\/sup>\u03b8 + 3 tan<sup>2&nbsp;<\/sup>\u03b8 sec<sup>2&nbsp;<\/sup>\u03b8 + 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:&nbsp;<\/strong><\/p>\n\n\n\n<p>From trig. Identities we have,<\/p>\n\n\n\n<p>sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 tan<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>On cubing&nbsp;both sides,<\/p>\n\n\n\n<p>(sec<sup>2<\/sup>\u03b8 \u2212 tan<sup>2<\/sup>\u03b8)<sup>3&nbsp;<\/sup>= 1<\/p>\n\n\n\n<p>sec<sup>6&nbsp;<\/sup>\u03b8 \u2212 tan<sup>6&nbsp;<\/sup>\u03b8 \u2212 3sec<sup>2&nbsp;<\/sup>\u03b8 tan<sup>2&nbsp;<\/sup>\u03b8(sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 tan<sup>2&nbsp;<\/sup>\u03b8) = 1[Since,&nbsp;(a&nbsp;\u2013 b)<sup>3<\/sup>&nbsp;= a<sup>3<\/sup>&nbsp;\u2013 b<sup>3<\/sup>&nbsp;\u2013 3ab(a \u2013 b)]<\/p>\n\n\n\n<p>sec<sup>6&nbsp;<\/sup>\u03b8 \u2212 tan<sup>6&nbsp;<\/sup>\u03b8 \u2212 3sec<sup>2&nbsp;<\/sup>\u03b8 tan<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>\u21d2 sec<sup>6&nbsp;<\/sup>\u03b8 = tan<sup>6&nbsp;<\/sup>\u03b8 + 3sec<sup>2&nbsp;<\/sup>\u03b8 tan<sup>2&nbsp;<\/sup>\u03b8 + 1<\/p>\n\n\n\n<p>Hence, L.H.S = R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>32. cosec<sup>6&nbsp;<\/sup>\u03b8 = cot<sup>6&nbsp;<\/sup>\u03b8 + 3cot<sup>2&nbsp;<\/sup>\u03b8 cosec<sup>2&nbsp;<\/sup>\u03b8 + 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>From trig. Identities we have,<\/p>\n\n\n\n<p>cosec<sup>2&nbsp;<\/sup>\u03b8 \u2212 cot<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>On cubing&nbsp;both sides,<\/p>\n\n\n\n<p>(cosec<sup>2&nbsp;<\/sup>\u03b8 \u2212 cot<sup>2&nbsp;<\/sup>\u03b8)<sup>3<\/sup>&nbsp;= 1<\/p>\n\n\n\n<p>cosec<sup>6&nbsp;<\/sup>\u03b8 \u2212 cot<sup>6&nbsp;<\/sup>\u03b8 \u2212 3cosec<sup>2&nbsp;<\/sup>\u03b8 cot<sup>2&nbsp;<\/sup>\u03b8 (cosec<sup>2&nbsp;<\/sup>\u03b8 \u2212 cot<sup>2&nbsp;<\/sup>\u03b8) = 1[Since,&nbsp;(a&nbsp;\u2013 b)<sup>3<\/sup>&nbsp;= a<sup>3<\/sup>&nbsp;\u2013 b<sup>3<\/sup>&nbsp;\u2013 3ab(a \u2013 b)]<\/p>\n\n\n\n<p>cosec<sup>6&nbsp;<\/sup>\u03b8 \u2212 cot<sup>6&nbsp;<\/sup>\u03b8 \u2212 3cosec<sup>2&nbsp;<\/sup>\u03b8 cot<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>\u21d2 cosec<sup>6&nbsp;<\/sup>\u03b8 = cot<sup>6&nbsp;<\/sup>\u03b8 + 3 cosec<sup>2&nbsp;<\/sup>\u03b8 cot<sup>2&nbsp;<\/sup>\u03b8 + 1<\/p>\n\n\n\n<p>Hence, L.H.S = R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>33.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S and using sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 tan<sup>2&nbsp;<\/sup>\u03b8 = 1&nbsp;\u21d2 1 + tan<sup>2&nbsp;<\/sup>\u03b8 = sec<sup>2&nbsp;<\/sup>\u03b8<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"188\" height=\"170\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-34.png\" alt=\"\" class=\"wp-image-544704\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>34.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"182\" height=\"55\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-35.png\" alt=\"\" class=\"wp-image-544705\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S and using the identity sin<sup>2<\/sup>A + cos<sup>2<\/sup>A&nbsp;= 1, we get<\/p>\n\n\n\n<p>sin<sup>2<\/sup>A = 1 \u2212 cos<sup>2<\/sup>A<\/p>\n\n\n\n<p>\u21d2 sin<sup>2<\/sup>A = (1 \u2013 cos A)(1 + cos A)<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"235\" height=\"111\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-36.png\" alt=\"\" class=\"wp-image-544706\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>35.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We have,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"165\" height=\"51\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-38.png\" alt=\"\" class=\"wp-image-544707\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>Rationalizing the denominator and numerator with (sec A + tan A) and using sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 tan<sup>2&nbsp;<\/sup>\u03b8 = 1&nbsp;we get,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"273\" height=\"388\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-39.png\" alt=\"\" class=\"wp-image-544708\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-39.png 273w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-39-211x300.png 211w\" sizes=\"auto, (max-width: 273px) 100vw, 273px\" \/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"189\" height=\"54\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-40.png\" alt=\"\" class=\"wp-image-544709\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p><strong>36.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We have,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"153\" height=\"54\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-41.png\" alt=\"\" class=\"wp-image-544710\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-41.png 153w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-41-150x54.png 150w\" sizes=\"auto, (max-width: 153px) 100vw, 153px\" \/><\/figure>\n\n\n\n<p>On multiplying numerator and denominator by (1 \u2013 cos A), we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"209\" height=\"257\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-42.png\" alt=\"\" class=\"wp-image-544711\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"219\" height=\"71\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-43.png\" alt=\"\" class=\"wp-image-544712\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p><strong>37. (i)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S and rationalizing the numerator and denominator with \u221a(1 + sin A), we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"338\" height=\"228\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-44.png\" alt=\"\" class=\"wp-image-544713\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-44.png 338w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-44-300x202.png 300w\" sizes=\"auto, (max-width: 338px) 100vw, 338px\" \/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"312\" height=\"71\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-45.png\" alt=\"\" class=\"wp-image-544714\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-45.png 312w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-45-300x68.png 300w\" sizes=\"auto, (max-width: 312px) 100vw, 312px\" \/><\/figure>\n\n\n\n<p><strong>(ii)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"430\" height=\"426\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-46.png\" alt=\"\" class=\"wp-image-544715\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-46.png 430w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-46-300x297.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-46-150x150.png 150w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-46-400x396.png 400w\" sizes=\"auto, (max-width: 430px) 100vw, 430px\" \/><\/figure>\n\n\n\n<p>= 2 cosec A<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>38. Prove that:<\/strong><\/p>\n\n\n\n<p><strong>(i)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"420\" height=\"480\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-48.png\" alt=\"\" class=\"wp-image-544716\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-48.png 420w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-48-263x300.png 263w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-48-400x457.png 400w\" sizes=\"auto, (max-width: 420px) 100vw, 420px\" \/><\/figure>\n\n\n\n<p>= 2 cosec \u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>(ii)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"409\" height=\"447\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-50.png\" alt=\"\" class=\"wp-image-544717\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-50.png 409w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-50-274x300.png 274w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-50-400x437.png 400w\" sizes=\"auto, (max-width: 409px) 100vw, 409px\" \/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>(iii)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"413\" height=\"432\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-52.png\" alt=\"\" class=\"wp-image-544718\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-52.png 413w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-52-287x300.png 287w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-52-400x418.png 400w\" sizes=\"auto, (max-width: 413px) 100vw, 413px\" \/><\/figure>\n\n\n\n<p>= 2 cosec \u03b8<\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>(iv)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S, we have<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"390\" height=\"333\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-54.png\" alt=\"\" class=\"wp-image-544719\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-54.png 390w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-54-300x256.png 300w\" sizes=\"auto, (max-width: 390px) 100vw, 390px\" \/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"233\" height=\"50\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-55.png\" alt=\"\" class=\"wp-image-544720\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p><strong>39.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking LHS = (sec A \u2013 tan A)<sup>2<\/sup>&nbsp;, we have<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"200\" height=\"304\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-56.png\" alt=\"\" class=\"wp-image-544721\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-56.png 200w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-56-197x300.png 197w\" sizes=\"auto, (max-width: 200px) 100vw, 200px\" \/><\/figure>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"241\" height=\"56\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-57.png\" alt=\"\" class=\"wp-image-544722\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p><strong>40.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking L.H.S and rationalizing the numerator and denominator with (1 \u2013 cos A), we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"200\" height=\"260\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-58.png\" alt=\"\" class=\"wp-image-544723\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= (cosec A \u2013 cot A)<sup>2<\/sup><\/p>\n\n\n\n<p>= (cot A \u2013 cosec A)<sup>2<\/sup><\/p>\n\n\n\n<p>= R.H.S<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"314\" height=\"55\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-59.png\" alt=\"\" class=\"wp-image-544724\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-59.png 314w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-59-300x53.png 300w\" sizes=\"auto, (max-width: 314px) 100vw, 314px\" \/><\/figure>\n\n\n\n<p><strong>41.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Considering L.H.S and taking L.C.M and on simplifying we have,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"201\" height=\"364\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-60.png\" alt=\"\" class=\"wp-image-544725\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-60.png 201w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-60-166x300.png 166w\" sizes=\"auto, (max-width: 201px) 100vw, 201px\" \/><\/figure>\n\n\n\n<p>= 2 cosec A cot A = RHS<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"305\" height=\"47\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-61.png\" alt=\"\" class=\"wp-image-544726\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-61.png 305w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-61-300x47.png 300w\" sizes=\"auto, (max-width: 305px) 100vw, 305px\" \/><\/figure>\n\n\n\n<p><strong>42.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Taking LHS, we have<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"263\" height=\"310\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-62.png\" alt=\"\" class=\"wp-image-544727\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-62.png 263w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-62-255x300.png 255w\" sizes=\"auto, (max-width: 263px) 100vw, 263px\" \/><\/figure>\n\n\n\n<p>= cos A + sin A<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<p><strong>43.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Considering L.H.S and taking L.C.M and on simplifying we have,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"312\" height=\"243\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-64.png\" alt=\"\" class=\"wp-image-544728\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-64.png 312w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-64-300x234.png 300w\" sizes=\"auto, (max-width: 312px) 100vw, 312px\" \/><\/figure>\n\n\n\n<p>= 2 sec<sup>2&nbsp;<\/sup>A<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Hence proved<\/li><\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 6.2 Page No: 6.54<\/h4>\n\n\n\n<p><strong>1. If&nbsp;cos \u03b8 = 4\/5, find all other trigonometric ratios of angle&nbsp;\u03b8.&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We have,<\/p>\n\n\n\n<p>cos \u03b8 = 4\/5<\/p>\n\n\n\n<p>And we know that,<\/p>\n\n\n\n<p>sin \u03b8 = \u221a(1 \u2013 cos<sup>2&nbsp;<\/sup>\u03b8)<\/p>\n\n\n\n<p>\u21d2 sin \u03b8 = \u221a(1 \u2013 (4\/5)<sup>2<\/sup>)<\/p>\n\n\n\n<p>= \u221a(1 \u2013 (16\/25))<\/p>\n\n\n\n<p>= \u221a[(25 \u2013 16)\/25]<\/p>\n\n\n\n<p>= \u221a(9\/25)<\/p>\n\n\n\n<p>= 3\/5<\/p>\n\n\n\n<p>\u2234 sin \u03b8 = 3\/5<\/p>\n\n\n\n<p>Since, cosec \u03b8 = 1\/ sin \u03b8<\/p>\n\n\n\n<p>= 1\/ (3\/5)<\/p>\n\n\n\n<p>\u21d2 cosec \u03b8 = 5\/3<\/p>\n\n\n\n<p>And, sec \u03b8 = 1\/ cos \u03b8<\/p>\n\n\n\n<p>= 1\/ (4\/5)<\/p>\n\n\n\n<p>\u21d2 cosec \u03b8 = 5\/4<\/p>\n\n\n\n<p>Now,<\/p>\n\n\n\n<p>tan \u03b8 = sin \u03b8\/ cos \u03b8<\/p>\n\n\n\n<p>= (3\/5)\/ (4\/5)<\/p>\n\n\n\n<p>\u21d2 tan \u03b8 = 3\/4<\/p>\n\n\n\n<p>And, cot \u03b8 = 1\/ tan \u03b8<\/p>\n\n\n\n<p>= 1\/ (3\/4)<\/p>\n\n\n\n<p>\u21d2 cot \u03b8 = 4\/3<\/p>\n\n\n\n<p><strong>2. If&nbsp;sin \u03b8 = 1\/\u221a2, find all other trigonometric ratios of angle&nbsp;\u03b8.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We have,<\/p>\n\n\n\n<p>sin \u03b8 = 1\/\u221a2<\/p>\n\n\n\n<p>And we know that,<\/p>\n\n\n\n<p>cos \u03b8 = \u221a(1 \u2013 sin<sup>2&nbsp;<\/sup>\u03b8)<\/p>\n\n\n\n<p>\u21d2 cos \u03b8 = \u221a(1 \u2013 (1\/\u221a2)<sup>2<\/sup>)<\/p>\n\n\n\n<p>= \u221a(1 \u2013 (1\/2))<\/p>\n\n\n\n<p>= \u221a[(2 \u2013 1)\/2]<\/p>\n\n\n\n<p>= \u221a(1\/2)<\/p>\n\n\n\n<p>= 1\/\u221a2<\/p>\n\n\n\n<p>\u2234 cos \u03b8 = 1\/\u221a2<\/p>\n\n\n\n<p>Since, cosec \u03b8 = 1\/ sin \u03b8<\/p>\n\n\n\n<p>= 1\/ (1\/\u221a2)<\/p>\n\n\n\n<p>\u21d2 cosec \u03b8 = \u221a2<\/p>\n\n\n\n<p>And, sec \u03b8 = 1\/ cos \u03b8<\/p>\n\n\n\n<p>= 1\/ (1\/\u221a2)<\/p>\n\n\n\n<p>\u21d2 sec \u03b8 = \u221a2<\/p>\n\n\n\n<p>Now,<\/p>\n\n\n\n<p>tan \u03b8 = sin \u03b8\/ cos \u03b8<\/p>\n\n\n\n<p>= (1\/\u221a2)\/ (1\/\u221a2)<\/p>\n\n\n\n<p>\u21d2 tan \u03b8 = 1<\/p>\n\n\n\n<p>And, cot \u03b8 = 1\/ tan \u03b8<\/p>\n\n\n\n<p>= 1\/ (1)<\/p>\n\n\n\n<p>\u21d2 cot \u03b8 = 1<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"381\" height=\"56\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-65.png\" alt=\"\" class=\"wp-image-544729\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-65.png 381w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-65-300x44.png 300w\" sizes=\"auto, (max-width: 381px) 100vw, 381px\" \/><\/figure>\n\n\n\n<p><strong>3.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>tan \u03b8 = 1\/\u221a2<\/p>\n\n\n\n<p>By using sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 tan<sup>2&nbsp;<\/sup>\u03b8 = 1,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"443\" height=\"446\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-66.png\" alt=\"\" class=\"wp-image-544730\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-66.png 443w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-66-298x300.png 298w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-66-150x150.png 150w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-66-200x200.png 200w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-66-400x403.png 400w\" sizes=\"auto, (max-width: 443px) 100vw, 443px\" \/><\/figure>\n\n\n\n\n\n<p><strong>4.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>tan \u03b8 = 3\/4<\/p>\n\n\n\n<p>By using sec<sup>2&nbsp;<\/sup>\u03b8 \u2212 tan<sup>2&nbsp;<\/sup>\u03b8 = 1,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"577\" height=\"82\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-68.png\" alt=\"\" class=\"wp-image-544731\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-68.png 577w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-68-300x43.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-68-400x57.png 400w\" sizes=\"auto, (max-width: 577px) 100vw, 577px\" \/><\/figure>\n\n\n\n<p><strong>sec&nbsp;<\/strong>\u03b8 = 5\/4<\/p>\n\n\n\n<p>Since, sec \u03b8 = 1\/ cos \u03b8<\/p>\n\n\n\n<p>\u21d2 cos \u03b8 = 1\/ sec \u03b8<\/p>\n\n\n\n<p>= 1\/ (5\/4)<\/p>\n\n\n\n<p>= 4\/5<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"263\" height=\"77\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-69.png\" alt=\"\" class=\"wp-image-544732\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>So,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"312\" height=\"55\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-70.png\" alt=\"\" class=\"wp-image-544733\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-70.png 312w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-70-300x53.png 300w\" sizes=\"auto, (max-width: 312px) 100vw, 312px\" \/><\/figure>\n\n\n\n<p><strong>5.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given, tan \u03b8 = 12\/5<\/p>\n\n\n\n<p>Since, cot \u03b8 = 1\/ tan \u03b8 = 1\/ (12\/5) = 5\/12<\/p>\n\n\n\n<p>Now, by using cosec<sup>2&nbsp;<\/sup>\u03b8 \u2212 cot<sup>2&nbsp;<\/sup>\u03b8 = 1<\/p>\n\n\n\n<p>cosec \u03b8 = \u221a(1 + cot<sup>2&nbsp;<\/sup>\u03b8)<\/p>\n\n\n\n<p>= \u221a(1 + (5\/12)<sup>2&nbsp;<\/sup>)<\/p>\n\n\n\n<p>= \u221a(1 + 25\/144)<\/p>\n\n\n\n<p>= \u221a(169\/ 25)<\/p>\n\n\n\n<p>\u21d2 cosec \u03b8 = 13\/5<\/p>\n\n\n\n<p>Now, we know that<\/p>\n\n\n\n<p>sin \u03b8 = 1\/ cosec \u03b8<\/p>\n\n\n\n<p>= 1\/ (13\/5)<\/p>\n\n\n\n<p>\u21d2 sin \u03b8 = 5\/13<\/p>\n\n\n\n<p>Putting value of sin \u03b8 in the expression we have,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"212\" height=\"86\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-71.png\" alt=\"\" class=\"wp-image-544734\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= 25\/ 1<\/p>\n\n\n\n<p>= 25<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"323\" height=\"46\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-72.png\" alt=\"\" class=\"wp-image-544735\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-72.png 323w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-72-300x43.png 300w\" sizes=\"auto, (max-width: 323px) 100vw, 323px\" \/><\/figure>\n\n\n\n<p><strong>6.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Given,<\/strong><\/p>\n\n\n\n<p>cot \u03b8 = 1\/\u221a3<\/p>\n\n\n\n<p>Using cosec<sup>2&nbsp;<\/sup>\u03b8 \u2212 cot<sup>2&nbsp;<\/sup>\u03b8 = 1, we can find cosec \u03b8<\/p>\n\n\n\n<p>cosec \u03b8 = \u221a(1 + cot<sup>2<\/sup>&nbsp;\u03b8)<\/p>\n\n\n\n<p>= \u221a(1 + (1\/\u221a3)<sup>2<\/sup>)<\/p>\n\n\n\n<p>= \u221a(1 + (1\/3)) = \u221a((3 + 1)\/3)<\/p>\n\n\n\n<p>= \u221a(4\/3)<\/p>\n\n\n\n<p>\u21d2 cosec \u03b8 = 2\/\u221a3<\/p>\n\n\n\n<p>So, sin \u03b8 = 1\/ cosec \u03b8 = 1\/ (2\/\u221a3)<\/p>\n\n\n\n<p>\u21d2 sin \u03b8 = \u221a3\/2<\/p>\n\n\n\n<p>And, we know that<\/p>\n\n\n\n<p>cos \u03b8 = \u221a(1 \u2013 sin<sup>2<\/sup>&nbsp;\u03b8)<\/p>\n\n\n\n<p>= \u221a(1 \u2013 (\u221a3\/2)<sup>2<\/sup>)<\/p>\n\n\n\n<p>= \u221a(1 \u2013 (3\/4))<\/p>\n\n\n\n<p><strong>=&nbsp;<\/strong>\u221a((4 \u2013 3)\/4)<\/p>\n\n\n\n<p>= \u221a(1\/4)<\/p>\n\n\n\n<p>\u21d2 cos \u03b8 = 1\/2<\/p>\n\n\n\n<p>Now, using cos \u03b8 and sin \u03b8 in the expression, we have<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"158\" height=\"190\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-73.png\" alt=\"\" class=\"wp-image-544736\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\"\/><\/figure>\n\n\n\n<p>= 3\/5<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"411\" height=\"53\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-74.png\" alt=\"\" class=\"wp-image-544737\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-74.png 411w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-74-300x39.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-74-400x52.png 400w\" sizes=\"auto, (max-width: 411px) 100vw, 411px\" \/><\/figure>\n\n\n\n<p><strong>7.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>cosec A = \u221a2<\/p>\n\n\n\n<p>Using cosec<sup>2&nbsp;<\/sup>A \u2212 cot<sup>2&nbsp;<\/sup>A = 1, we find cot A<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"427\" height=\"498\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-75.png\" alt=\"\" class=\"wp-image-544738\" title=\"RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-75.png 427w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-75-257x300.png 257w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-6-75-400x467.png 400w\" sizes=\"auto, (max-width: 427px) 100vw, 427px\" \/><\/figure>\n\n\n\n<p>= 4\/2<\/p>\n\n\n\n<p>= 2<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-rd-sharma-solutions-for-class-10-maths-chapter-6-download-pdf\">RD Sharma Solutions for Class 10 Maths Chapter 6:&nbsp;<strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>RD Sharma Solutions for Class 10 Maths Chapter 6\u2013Trigonometric Identities<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RD-Sharma-Solutions-for-Class-10-Maths-Chapter-6\u2013Trigonometric-Identities.pdf\" target=\"_blank\" rel=\"noreferrer noopener\"><strong>Download PDF<\/strong>: RD Sharma Solutions for Class 10 Maths Chapter 6\u2013Trigonometric Identities PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Chapterwise RD Sharma Solutions for Class 10&nbsp;Maths :<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-1-real-numbers\/\">Chapter 1\u2013Real Numbers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\/\">Chapter 2\u2013Polynomials<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables\/\">Chapter 3\u2013Pair of Linear Equations In Two Variables<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-4-triangles\/\">Chapter 4\u2013Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-5-trigonometric-ratios\/\">Chapter 5\u2013Trigonometric Ratios<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-6-trigonometric-identities\/\">Chapter 6\u2013Trigonometric Identities<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-7-statistics\/\">Chapter 7\u2013Statistics<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-8-quadratic-equations\/\">Chapter 8\u2013Quadratic Equations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-9-arithmetic-progressions\/\">Chapter 9\u2013Arithmetic Progressions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-10-circles\/\">Chapter 10\u2013Circles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-11-constructions\/\">Chapter 11\u2013Constructions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry\/\">Chapter 12\u2013Some Applications of Trigonometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-13-probability\/\">Chapter 13\u2013Probability<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-14-co-ordinate-geometry\/\">Chapter 14\u2013Co-ordinate Geometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-15-areas-related-to-circles\/\">Chapter 15\u2013Areas Related To Circles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-16-surface-areas-and-volumes\/\">Chapter 16\u2013Surface Areas And Volumes<\/a><\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">About RD Sharma<\/h2>\n\n\n\n<p>RD Sharma i<em>sn&#8217;t the kind of author you&#8217;d bump into at lit fests. But his bestselling books have helped many&nbsp;<\/em>CBSE<em>&nbsp;students lose their dread of&nbsp;<\/em>maths<em>. Sunday Times profiles the tutor turned internet star<\/em><br>He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like &#8216;series solution of linear differential equations&#8217;. Meet Dr&nbsp;Ravi Dutt Sharma&nbsp;\u2014&nbsp;mathematics&nbsp;teacher and author of 25 reference books \u2014 whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it&#8217;s only recently that a spoof video turned the tutor into a YouTube star.<\/p>\n\n\n\n<p>R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. &#8220;I like to spend all my time thinking and writing about maths problems. I find it relaxing,&#8221; he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government&#8217;s Guru Nanak Dev Institute of Technology.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Read More<\/h2>\n\n\n\n<ul class=\"wp-block-yoast-seo-related-links\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-8-introduction-to-trigonometry\/\">NCERT Solutions for Class 10 Maths Chapter 8 &#8211; Introduction to Trigonometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-7-values-of-trigonometric-functions-at-sum-or-difference-of-angles\/\">RD Sharma Solutions for Class 11 Maths Chapter 7\u2013Values of Trigonometric Functions at Sum or Difference of Angles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-12-maths-chapter-4-inverse-trigonometric-functions\/\">RD Sharma Solutions for Class 12 Maths Chapter 4\u2013Inverse Trigonometric Functions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-5-trigonometric-functions\/\">RD Sharma Solutions for Class 11 Maths Chapter 5\u2013Trigonometric Functions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-9-values-of-trigonometric-functions-at-multiples-and-submultiples-of-an-angle\/\">RD Sharma Solutions for Class 11 Maths Chapter 9\u2013Values of Trigonometric Functions at Multiples and Submultiples of an Angle<\/a><\/li><\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Class 10: Maths Chapter 6 solutions. Complete Class 10 Maths Chapter 6 Notes. RD Sharma Solutions for Class 10 Maths Chapter 6\u2013Trigonometric Identities Class 10: Maths Chapter 6 solutions. Complete Class 10 Maths Chapter 6 Notes. Exercise 6.1 Page No: 6.43 Prove the following trigonometric identities: 1. (1 \u2013 cos2&nbsp;A) cosec2&nbsp;A = 1 Solution: Taking [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":544675,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,24],"tags":[1962],"boards":[],"class_list":["post-544672","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-10","tag-rd-sharma-solutions","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>RD Sharma Solutions for Class 10, maths Chapter 6 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"RD Sharma Solutions for Class 10 Maths Chapter 6\u2013Trigonometric Identities | Browse Class 10 Maths Chapters RD Sharma books - IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-6-trigonometric-identities\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"RD Sharma Solutions for Class 10 Maths Chapter 6\u2013Trigonometric Identities\" \/>\n<meta property=\"og:description\" content=\"Class 10: Maths Chapter 6 solutions. Complete Class 10 Maths Chapter 6 Notes. 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