{"id":67932,"date":"2023-08-31T14:09:00","date_gmt":"2023-08-31T08:39:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67932"},"modified":"2023-09-11T14:58:28","modified_gmt":"2023-09-11T09:28:28","slug":"rd-sharma-solutions-class-11-maths-chapter-9-exercise-9-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-9-exercise-9-1\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 9 Exercise 9.1 (Updated  for 2023-24)"},"content":{"rendered":"\n<p><img class=\"alignnone wp-image-122273 size-full\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-9-Exercise-9.1.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-9-Exercise-9.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-9-Exercise-9.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1:<\/strong> If you are looking for a good guidebook for your Class 11 Maths, then we would recommend <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths<\/a>. You can prepare for your exams or make your assignments with this without any hassle. To know more about the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-9-trigonometric-ratios-of-multiple-and-sub-multiple-angles\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 9<\/a> Exercise 9.1, read the complete blog.&nbsp;<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69e7673e663d9\" 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-69e7673e663d9\"><\/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-9-exercise-9-1\/#download-rd-sharma-solutions-class-11-maths-chapter-9-exercise-91-pdf\" title=\"Download RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1 PDF:\">Download RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1 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-9-exercise-9-1\/#access-answers-of-rd-sharma-solutions-class-11-maths-chapter-9-exercise-91\" title=\"Access answers of RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1\">Access answers of RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-9-exercise-9-1\/#faqs-of-rd-sharma-solutions-class-11-maths-chapter-11\" title=\"FAQs of RD Sharma Solutions Class 11 Maths Chapter 11\">FAQs of RD Sharma Solutions Class 11 Maths Chapter 11<\/a><ul class='ez-toc-list-level-3'><li class='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-9-exercise-9-1\/#how-many-questions-are-there-in-rd-sharma-solutions-for-class-11-maths-chapter-9-exercise-91\" title=\"How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.1?\">How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.1?<\/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-9-exercise-9-1\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-9-exercise-91\" title=\"From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1?\">From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1?<\/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-9-exercise-9-1\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-11-maths-chapter-9-exercise-91\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.1?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.1?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-9-exercise-91-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1 PDF:<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-9-Ex-9.1-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 11 Solutions Chapter 9 Exercise 9.1<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-9-Ex-9.1-1.pdf\",\"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-class-11-maths-chapter-9-exercise-91\"><\/span>Access answers of RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<article id=\"post-52886\" class=\"post-52886 page type-page status-publish hentry\">\n<p><strong>Prove the following identities:<\/strong><\/p>\n<p><strong>1. \u221a[(1 \u2013 cos 2x) \/ (1 + cos 2x)] = tan x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>\u221a[(1 \u2013 cos 2x) \/ (1 + cos 2x)]<\/p>\n<p>We know that cos 2x = 1 \u2013 2 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>= 2 cos<sup>2<\/sup>&nbsp;x \u2013 1<\/p>\n<p>So,<\/p>\n<p>\u221a[(1 \u2013 cos 2x) \/ (1 + cos 2x)] = \u221a[(1 \u2013 (1 \u2013 2sin<sup>2<\/sup>&nbsp;x)) \/ (1 + (2cos<sup>2<\/sup>x \u2013 1))]<\/p>\n<p>= \u221a[(1 \u2013 1 + 2sin<sup>2<\/sup>&nbsp;x) \/ (1 + 2cos<sup>2<\/sup>&nbsp;x \u2013 1)]<\/p>\n<p>= \u221a[2 sin<sup>2<\/sup>&nbsp;x \/ 2 cos<sup>2<\/sup>&nbsp;x]<\/p>\n<p>= sin x\/cos x<\/p>\n<p>= tan x<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>2. sin 2x \/ (1 \u2013 cos 2x) = cot x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>sin 2x \/ (1 \u2013 cos 2x)<\/p>\n<p>We know that cos 2x = 1 \u2013 2 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>Sin 2x = 2 sin x cos x<\/p>\n<p>So,<\/p>\n<p>sin 2x \/ (1 \u2013 cos 2x) = (2 sin x cos x) \/ (1 \u2013 (1 \u2013 2sin<sup>2<\/sup>&nbsp;x))<\/p>\n<p>= (2 sin x cos x) \/ (1 \u2013 1 + 2sin<sup>2<\/sup>&nbsp;x)]<\/p>\n<p>= [2 sin x cos x \/ 2 sin<sup>2<\/sup>&nbsp;x]<\/p>\n<p>= cos x\/sin x<\/p>\n<p>= cot x<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>3. sin 2x \/ (1 + cos 2x) = tan x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>sin 2x \/ (1 + cos 2x)<\/p>\n<p>We know that cos 2x = 1 \u2013 2 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>= 2 cos<sup>2<\/sup>&nbsp;x \u2013 1<\/p>\n<p>Sin 2x = 2 sin x cos x<\/p>\n<p>So,<\/p>\n<p>sin 2x \/ (1 + cos 2x) = [2 sin x cos x \/ (1 + (2cos<sup>2<\/sup>x \u2013 1))]<\/p>\n<p>= [2 sin x cos x \/ (1 + 2cos<sup>2<\/sup>&nbsp;x \u2013 1)]<\/p>\n<p>= [2 sin x cos x \/ 2 cos<sup>2<\/sup>&nbsp;x]<\/p>\n<p>= sin x\/cos x<\/p>\n<p>= tan x<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>5. [1 \u2013 cos 2x + sin 2x] \/ [1 + cos 2x + sin 2x] = tan x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n[1 \u2013 cos 2x + sin 2x] \/ [1 + cos 2x + sin 2x]\n<p>We know that, cos 2x = 1 \u2013 2 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>= 2 cos<sup>2<\/sup>&nbsp;x \u2013 1<\/p>\n<p>Sin 2x = 2 sin x cos x<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-1.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 2\"><\/p>\n<p><strong>6. [sin x + sin 2x] \/ [1 + cos x + cos 2x] = tan x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n[sin x + sin 2x] \/ [1 + cos x + cos 2x]\n<p>We know that, cos 2x = cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>Sin 2x = 2 sin x cos x<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-2.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 3\"><\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>7. cos 2x \/ (1 + sin 2x) = tan (\u03c0\/4 \u2013 x)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>cos 2x \/ (1 + sin 2x)<\/p>\n<p>We know that, cos 2x = cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>Sin 2x = 2 sin x cos x<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-3.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 4\"><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 5\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-4.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 5\"><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-5.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 6\"><\/p>\n<p><strong>8. cos x \/ (1 \u2013 sin x) = tan (\u03c0\/4 + x\/2)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>cos x \/ (1 \u2013 sin x)<\/p>\n<p>We know that, cos 2x = cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>Cos x = cos<sup>2<\/sup>&nbsp;x\/2 \u2013 sin<sup>2<\/sup>&nbsp;x\/2<\/p>\n<p>Sin 2x = 2 sin x cos x<\/p>\n<p>Sin x = 2 sin x\/2 cos x\/2<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 7\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-6.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 7\"><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 8\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-7.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 8\"><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 9\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-8.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 9\"><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 10\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-9.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 10\"><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 11\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-10.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 11\"><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 12\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-11.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 12\"><\/p>\n<p><strong>11. (cos \u03b1 + cos \u03b2)<sup>&nbsp;2<\/sup>&nbsp;+ (sin \u03b1 + sin \u03b2)<sup>&nbsp;2<\/sup>&nbsp;= 4 cos<sup>2<\/sup>&nbsp;(\u03b1 \u2013 \u03b2)\/2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>(cos \u03b1 + cos \u03b2)<sup>2<\/sup>&nbsp;+ (sin \u03b1 + sin \u03b2)<sup>2<\/sup><\/p>\n<p>Upon expansion, we get,<\/p>\n<p>(cos \u03b1 + cos \u03b2)<sup>2<\/sup>&nbsp;+ (sin \u03b1 + sin \u03b2)<sup>2<\/sup>&nbsp;=<\/p>\n<p>= cos<sup>2<\/sup>&nbsp;\u03b1 + cos<sup>2<\/sup>&nbsp;\u03b2 + 2 cos \u03b1 cos \u03b2 + sin<sup>2<\/sup>&nbsp;\u03b1 + sin<sup>2<\/sup>&nbsp;\u03b2 + 2 sin \u03b1 sin \u03b2<\/p>\n<p>= 2 + 2 cos \u03b1 cos \u03b2 + 2 sin \u03b1 sin \u03b2<\/p>\n<p>= 2 (1 + cos \u03b1 cos \u03b2 + sin \u03b1 sin \u03b2)<\/p>\n<p>= 2 (1 + cos (\u03b1 \u2013 \u03b2)) [since, cos (A \u2013 B) = cos A cos B + sin A sin B]<\/p>\n<p>= 2 (1 + 2 cos<sup>2<\/sup>&nbsp;(\u03b1 \u2013 \u03b2)\/2 \u2013 1) [since, cos2x = 2cos<sup>2<\/sup>&nbsp;x \u2013 1]<\/p>\n<p>= 2 (2 cos<sup>2<\/sup>&nbsp;(\u03b1 \u2013 \u03b2)\/2)<\/p>\n<p>= 4 cos<sup>2<\/sup>&nbsp;(\u03b1 \u2013 \u03b2)\/2<\/p>\n<p>= RHS<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>12. sin<sup>2<\/sup>&nbsp;(\u03c0\/8 + x\/2) \u2013 sin<sup>2<\/sup>&nbsp;(\u03c0\/8 \u2013 x\/2) = 1\/\u221a2 sin x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>sin<sup>2<\/sup>&nbsp;(\u03c0\/8 + x\/2) \u2013 sin<sup>2<\/sup>&nbsp;(\u03c0\/8 \u2013 x\/2)<\/p>\n<p>we know, sin<sup>2<\/sup>&nbsp;A \u2013 sin<sup>2<\/sup>&nbsp;B = sin (A+B) sin (A-B)<\/p>\n<p>so,<\/p>\n<p>sin<sup>2<\/sup>&nbsp;(\u03c0\/8 + x\/2) \u2013 sin<sup>2<\/sup>&nbsp;(\u03c0\/8 \u2013 x\/2) = sin (\u03c0\/8 + x\/2 + \u03c0\/8 \u2013 x\/2) sin (\u03c0\/8 + x\/2 \u2013 (\u03c0\/8 \u2013 x\/2))<\/p>\n<p>= sin (\u03c0\/8 + \u03c0\/8) sin (\u03c0\/8 + x\/2 \u2013 \u03c0\/8 + x\/2)<\/p>\n<p>= sin \u03c0\/4 sin x<\/p>\n<p>= 1\/\u221a2 sin x [since, since \u03c0\/4 = 1\/\u221a2]<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>13. 1 + cos<sup>2<\/sup>&nbsp;2x = 2 (cos<sup>4<\/sup>&nbsp;x + sin<sup>4<\/sup>&nbsp;x)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>1 + cos<sup>2<\/sup>&nbsp;2x<\/p>\n<p>We know, cos2x = cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>cos<sup>2<\/sup>&nbsp;x + sin<sup>2<\/sup>&nbsp;x = 1<\/p>\n<p>so,<\/p>\n<p>1 + cos<sup>2<\/sup>&nbsp;2x = (cos<sup>2<\/sup>&nbsp;x + sin<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>&nbsp;+ (cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup><\/p>\n<p>= (cos<sup>4<\/sup>&nbsp;x + sin<sup>4<\/sup>&nbsp;x + 2 cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x) + (cos<sup>4<\/sup>&nbsp;x + sin<sup>4<\/sup>&nbsp;x \u2013 2 cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x)<\/p>\n<p>= cos<sup>4<\/sup>&nbsp;x + sin<sup>4<\/sup>&nbsp;x + cos<sup>4<\/sup>&nbsp;x + sin<sup>4<\/sup>&nbsp;x<\/p>\n<p>= 2 cos<sup>4<\/sup>&nbsp;x + 2 sin<sup>4<\/sup>&nbsp;x<\/p>\n<p>= 2 (cos<sup>4<\/sup>&nbsp;x + sin<sup>4<\/sup>&nbsp;x)<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>14. cos<sup>3<\/sup>&nbsp;2x + 3 cos 2x = 4 (cos<sup>6<\/sup>&nbsp;x \u2013 sin<sup>6<\/sup>&nbsp;x)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider RHS:<\/p>\n<p>4 (cos<sup>6<\/sup>&nbsp;x \u2013 sin<sup>6<\/sup>&nbsp;x)<\/p>\n<p>Upon expansion, we get,<\/p>\n<p>4 (cos<sup>6<\/sup>&nbsp;x \u2013 sin<sup>6<\/sup>&nbsp;x) = 4 [(cos<sup>2<\/sup>&nbsp;x)<sup>3<\/sup>&nbsp;\u2013 (sin<sup>2<\/sup>&nbsp;x)<sup>3<\/sup>]<\/p>\n<p>= 4 (cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x) (cos<sup>4<\/sup>&nbsp;x + sin<sup>4<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x)<\/p>\n<p>By using the formula,<\/p>\n<p>a<sup>3<\/sup>&nbsp;\u2013 b<sup>3<\/sup>&nbsp;= (a-b) (a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>&nbsp;+ ab)<\/p>\n<p>= 4 cos 2x (cos<sup>4<\/sup>&nbsp;x + sin<sup>4<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x \u2013 cos<sup>2<\/sup>&nbsp;x sin<\/p>\n<p>We know, cos 2x = cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>So,<\/p>\n<p>= 4 cos 2x (cos<sup>4<\/sup>&nbsp;x + sin<sup>4<\/sup>&nbsp;x + 2 cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x \u2013 cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x)<\/p>\n<p>= 4 cos 2x [(cos<sup>2<\/sup>&nbsp;x)<sup>2<\/sup>&nbsp;+ (sin<sup>2<\/sup>&nbsp;x)<sup>2<\/sup>&nbsp;+ 2 cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x \u2013 cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x]<\/p>\n<p>We know, a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>&nbsp;+ 2ab = (a + b)<sup>2<\/sup><\/p>\n<p>= 4 cos 2x [(1)<sup>2<\/sup>&nbsp;\u2013 1\/4 (4 cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x)]<\/p>\n<p>= 4 cos 2x [(1)<sup>2<\/sup>&nbsp;\u2013 1\/4 (2 cos x sin x)<sup>2<\/sup>]<\/p>\n<p>We know, sin 2x = 2sin x cos x<\/p>\n<p>= 4 cos 2x [(1<sup>2<\/sup>) \u2013 1\/4 (sin 2x)<sup>2<\/sup>]<\/p>\n<p>= 4 cos 2x (1 \u2013 1\/4 sin<sup>2<\/sup>&nbsp;2x)<\/p>\n<p>We know, sin<sup>2<\/sup>&nbsp;x = 1 \u2013 cos<sup>2<\/sup>&nbsp;x<\/p>\n<p>= 4 cos 2x [1 \u2013 1\/4 (1 \u2013 cos<sup>2<\/sup>&nbsp;2x)]<\/p>\n<p>= 4 cos 2x [1 \u2013 1\/4 + 1\/4 cos<sup>2<\/sup>&nbsp;2x]<\/p>\n<p>= 4 cos 2x [3\/4 + 1\/4 cos<sup>2<\/sup>&nbsp;2x]<\/p>\n<p>= 4 (3\/4 cos 2x + 1\/4 cos<sup>3<\/sup>&nbsp;2x)<\/p>\n<p>= 3 cos 2x + cos<sup>3<\/sup>&nbsp;2x<\/p>\n<p>= cos<sup>3<\/sup>&nbsp;2x + 3 cos 2x<\/p>\n<p>= LHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>15. (sin 3x + sin x) sin x + (cos 3x \u2013 cos x) cos x = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>(sin 3x + sin x) sin x + (cos 3x \u2013 cos x) cos x<\/p>\n<p>= (sin 3x) (sin x) + sin<sup>2<\/sup>&nbsp;x + (cos 3x) (cos x) \u2013 cos<sup>2<\/sup>&nbsp;x<\/p>\n<p>= [(sin 3x) (sin x) + (cos 3x) (cos x)] + (sin<sup>2<\/sup>&nbsp;x \u2013 cos<sup>2<\/sup>&nbsp;x)<\/p>\n<p>= [(sin 3x) (sin x) + (cos 3x) (cos x)] \u2013 (cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x)<\/p>\n<p>= cos (3x \u2013 x) \u2013 cos 2x<\/p>\n<p>We know,&nbsp;cos 2x =&nbsp;cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>cos A cos B + sin A sin B = cos(A \u2013 B)<\/p>\n<p>So,<\/p>\n<p>= cos 2x \u2013 cos 2x<\/p>\n<p>= 0<\/p>\n<p>= RHS<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>16. cos<sup>2<\/sup>&nbsp;(\u03c0\/4 \u2013 x) \u2013 sin<sup>2<\/sup>&nbsp;(\u03c0\/4 \u2013 x) = sin 2x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>cos<sup>2<\/sup>&nbsp;(\u03c0\/4 \u2013 x) \u2013 sin<sup>2<\/sup>&nbsp;(\u03c0\/4 \u2013 x)<\/p>\n<p>We know, cos<sup>2<\/sup>&nbsp;A \u2013 sin<sup>2<\/sup>&nbsp;A = cos 2A<\/p>\n<p>So,<\/p>\n<p>cos<sup>2<\/sup>&nbsp;(\u03c0\/4 \u2013 x) \u2013 sin<sup>2<\/sup>&nbsp;(\u03c0\/4 \u2013 x) = cos 2 (\u03c0\/4 \u2013 x)<\/p>\n<p>= cos (\u03c0\/2 \u2013 2x)<\/p>\n<p>= sin 2x [since, cos (\u03c0\/2 \u2013 A) = sin A]<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>17. cos 4x = 1 \u2013 8 cos<sup>2<\/sup>&nbsp;x + 8 cos<sup>4<\/sup>&nbsp;x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>cos 4x<\/p>\n<p>We know, cos 2x = 2 cos<sup>2<\/sup>&nbsp;x \u2013 1<\/p>\n<p>So,<\/p>\n<p>cos 4x = 2 cos<sup>2<\/sup>&nbsp;2x \u2013 1<\/p>\n<p>= 2(2 cos<sup>2<\/sup>&nbsp;2x \u2013 1)<sup>2<\/sup>&nbsp;\u2013 1<\/p>\n<p>= 2[(2 cos<sup>2<\/sup>&nbsp;2x)<sup>&nbsp;2<\/sup>&nbsp;+ 1<sup>2<\/sup>&nbsp;\u2013 2\u00d72 cos<sup>2<\/sup>&nbsp;x] \u2013 1<\/p>\n<p>= 2(4 cos<sup>4<\/sup>&nbsp;2x + 1 \u2013 4 cos<sup>2<\/sup>&nbsp;x) \u2013 1<\/p>\n<p>= 8 cos<sup>4<\/sup>&nbsp;2x + 2 \u2013 8 cos<sup>2<\/sup>&nbsp;x \u2013 1<\/p>\n<p>= 8 cos<sup>4<\/sup>&nbsp;2x + 1 \u2013 8 cos<sup>2<\/sup>&nbsp;x<\/p>\n<p>= RHS<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>18. sin 4x = 4 sin x cos<sup>3<\/sup>&nbsp;x \u2013 4 cos x sin<sup>3<\/sup>&nbsp;x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>sin 4x<\/p>\n<p>We know, sin 2x = 2 sin x cos x<\/p>\n<p>cos 2x = cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>So,<\/p>\n<p>sin 4x = 2 sin 2x cos 2x<\/p>\n<p>= 2 (2 sin x cos x) (cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x)<\/p>\n<p>= 4 sin x cos x (cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x)<\/p>\n<p>= 4 sin x cos<sup>3<\/sup>&nbsp;x \u2013 4 sin<sup>3<\/sup>&nbsp;x cos x<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>19. 3(sin x \u2013 cos x)<sup>&nbsp;4<\/sup>&nbsp;+ 6 (sin x + cos x)<sup>&nbsp;2<\/sup>&nbsp;+ 4 (sin<sup>6<\/sup>&nbsp;x + cos<sup>6<\/sup>&nbsp;x) = 13<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>3(sin x \u2013 cos x)<sup>&nbsp;4<\/sup>&nbsp;+ 6 (sin x + cos x)<sup>&nbsp;2<\/sup>&nbsp;+ 4 (sin<sup>6<\/sup>&nbsp;x + cos<sup>6<\/sup>&nbsp;x)<\/p>\n<p>We know, (a + b)<sup>2<\/sup>&nbsp;= a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>&nbsp;+ 2ab<\/p>\n<p>(a \u2013 b)<sup>2<\/sup>&nbsp;= a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>&nbsp;\u2013 2ab<\/p>\n<p>a<sup>3<\/sup>&nbsp;+ b<sup>3<\/sup>&nbsp;= (a + b) (a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>&nbsp;\u2013 ab)<\/p>\n<p>So,<\/p>\n<p>3(sin x \u2013 cos x)<sup>&nbsp;4<\/sup>&nbsp;+ 6 (sin x + cos x)<sup>&nbsp;2<\/sup>&nbsp;+ 4 (sin<sup>6<\/sup>&nbsp;x + cos<sup>6<\/sup>&nbsp;x) = 3{(sin x \u2013 cos x)<sup>&nbsp;2<\/sup>}<sup>2<\/sup>&nbsp;+ 6 {(sin x)<sup>2<\/sup>&nbsp;+ (cos x)<sup>2<\/sup>&nbsp;+ 2 sin x cos x)} + 4 {(sin<sup>2<\/sup>&nbsp;x)<sup>3<\/sup>&nbsp;+ (cos<sup>2<\/sup>&nbsp;x)<sup>3<\/sup>}<\/p>\n<p>= 3{(sin x)<sup>&nbsp;2<\/sup>&nbsp;+ (cos x)<sup>2<\/sup>&nbsp;\u2013 2 sin x cos x)}<sup>2<\/sup>&nbsp;+ 6 (sin<sup>2<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x + 2 sin x cos x) + 4{(sin<sup>2<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x) (sin<sup>4<\/sup>&nbsp;x + cos<sup>4<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x)}<\/p>\n<p>= 3(1 \u2013 2 sin x cos x)&nbsp;<sup>2<\/sup>&nbsp;+ 6 (1 + 2 sin x cos x) + 4{(1) (sin<sup>4<\/sup>&nbsp;x + cos<sup>4<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x)}<\/p>\n<p>We know,&nbsp;sin<sup>2<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x = 1<\/p>\n<p>So,<\/p>\n<p>= 3{1<sup>2<\/sup>&nbsp;+ (2 sin x cos x)<sup>&nbsp;2<\/sup>&nbsp;\u2013 4 sin x cos x} + 6 (1 + 2 sin x cos x) + 4{(sin<sup>2<\/sup>&nbsp;x)<sup>2<\/sup>&nbsp;+ (cos<sup>2<\/sup>&nbsp;x)<sup>2<\/sup>&nbsp;+ 2 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x \u2013 3 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x)}<\/p>\n<p>= 3{1 + 4 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x \u2013 4 sin x cos x} + 6 (1 + 2 sin x cos x) + 4{(sin<sup>2<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>&nbsp;\u2013 3 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x)}<\/p>\n<p>= 3 + 12 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x \u2013 12 sin x cos x + 6 + 12 sin x cos x + 4{(1)<sup>2<\/sup>&nbsp;\u2013 3 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x)}<\/p>\n<p>= 9 + 12 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x + 4(1 \u2013 3 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x)<\/p>\n<p>= 9 + 12 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x + 4 \u2013 12 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x<\/p>\n<p>= 13<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>20. 2(sin<sup>6<\/sup>&nbsp;x + cos<sup>6<\/sup>&nbsp;x) \u2013 3(sin<sup>4<\/sup>&nbsp;x + cos<sup>4<\/sup>&nbsp;x) + 1 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>2(sin<sup>6<\/sup>&nbsp;x + cos<sup>6<\/sup>&nbsp;x) \u2013 3(sin<sup>4<\/sup>&nbsp;x + cos<sup>4<\/sup>&nbsp;x) + 1<\/p>\n<p>We know, (a + b)<sup>2<\/sup>&nbsp;= a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>&nbsp;+ 2ab<\/p>\n<p>a<sup>3<\/sup>&nbsp;+ b<sup>3<\/sup>&nbsp;= (a + b) (a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>&nbsp;\u2013 ab)<\/p>\n<p>So,<\/p>\n<p>2(sin<sup>6<\/sup>&nbsp;x + cos<sup>6<\/sup>&nbsp;x) \u2013 3(sin<sup>4<\/sup>&nbsp;x + cos<sup>4<\/sup>&nbsp;x) + 1 =&nbsp;2{(sin<sup>2<\/sup>&nbsp;x)<sup>&nbsp;3<\/sup>&nbsp;+ (cos<sup>2<\/sup>&nbsp;x)<sup>&nbsp;3<\/sup>} \u2013 3{(sin<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>&nbsp;+ (cos<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>} + 1<\/p>\n<p>=&nbsp;2{(sin<sup>2<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x) (sin<sup>4<\/sup>&nbsp;x + cos<sup>4<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x} \u2013 3{(sin<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>&nbsp;+ (cos<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>&nbsp;+ 2sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x \u2013 2sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x} + 1<\/p>\n<p>=&nbsp;2{(1) (sin<sup>4<\/sup>&nbsp;x + cos<sup>4<\/sup>&nbsp;x + 2 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x \u2013 3 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x} \u2013 3{(sin<sup>2<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>&nbsp;\u2013 2sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x} + 1<\/p>\n<p>&nbsp;<\/p>\n<p>We know, sin<sup>2<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x = 1<\/p>\n<p>=&nbsp;2{(sin<sup>2<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>&nbsp;\u2013 3 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x} \u2013 3{(1)<sup>2<\/sup>&nbsp;\u2013 2sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x} + 1<\/p>\n<p>=&nbsp;2{(1)<sup>2<\/sup>&nbsp;\u2013 3 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x} \u2013 3(1 \u2013 2sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x) + 1<\/p>\n<p>=&nbsp;2(1 \u2013 3 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x) \u2013 3 + 6 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x + 1<\/p>\n<p>=&nbsp;2 \u2013 6 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x \u2013 2 + 6 sin<sup>2<\/sup>&nbsp;x cos<sup>2<\/sup>&nbsp;x<\/p>\n<p>= 0<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>21. cos<sup>6<\/sup>&nbsp;x \u2013 sin<sup>6<\/sup>&nbsp;x = cos 2x (1 \u2013 1\/4 sin<sup>2<\/sup>&nbsp;2x)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>cos<sup>6<\/sup>&nbsp;x \u2013 sin<sup>6<\/sup>&nbsp;x<\/p>\n<p>We know, (a + b)<sup>&nbsp;2<\/sup>&nbsp;= a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>&nbsp;+ 2ab<\/p>\n<p>a<sup>3<\/sup>&nbsp;\u2013 b<sup>3<\/sup>&nbsp;= (a \u2013 b) (a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>&nbsp;+ ab)<\/p>\n<p>So,<\/p>\n<p>cos<sup>6<\/sup>&nbsp;x \u2013 sin<sup>6<\/sup>&nbsp;x = (cos<sup>2<\/sup>&nbsp;x)<sup>3<\/sup>&nbsp;\u2013 (sin<sup>2<\/sup>&nbsp;x)<sup>3<\/sup><\/p>\n<p>= (cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x) (cos<sup>4<\/sup>&nbsp;x + sin<sup>4<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x)<\/p>\n<p>We know, cos 2x = cos<sup>2<\/sup>&nbsp;x \u2013 sin<sup>2<\/sup>&nbsp;x<\/p>\n<p>So,<\/p>\n<p>= cos 2x [(cos<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>&nbsp;+ (sin<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>&nbsp;+ 2 cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x \u2013 cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x]<\/p>\n<p>= cos 2x [(cos<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>&nbsp;+ (sin<sup>2<\/sup>&nbsp;x)<sup>&nbsp;2<\/sup>&nbsp;\u2013 1\/4 \u00d7 4 cos<sup>2<\/sup>&nbsp;x sin<sup>2<\/sup>&nbsp;x]<\/p>\n<p>We know, sin<sup>2<\/sup>&nbsp;x + cos<sup>2<\/sup>&nbsp;x = 1<\/p>\n<p>So,<\/p>\n<p>= cos 2x [(1)<sup>2<\/sup>&nbsp;\u2013 1\/4 \u00d7 (2 cos x sin x)<sup>&nbsp;2<\/sup>]<\/p>\n<p>We know, sin 2x = 2 sin x cos x<\/p>\n<p>So,<\/p>\n<p>= cos 2x [1 \u2013 1\/4 \u00d7 (sin 2x)<sup>&nbsp;2<\/sup>]<\/p>\n<p>= cos 2x [1 \u2013 1\/4 \u00d7 sin<sup>2<\/sup>&nbsp;2x]<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>22. tan (\u03c0\/4 + x) + tan (\u03c0\/4 \u2013 x) = 2 sec 2x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>tan (\u03c0\/4 + x) + tan (\u03c0\/4 \u2013 x)<\/p>\n<p>We know,<\/p>\n<p>tan (A+B) = (tan A + tan B)\/(1- tan A tan B)<\/p>\n<p>tan (A-B) = (tan A \u2013 tan B)\/(1+ tan A tan B)<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-12.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 13\"><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 14\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-13.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 14\"><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 15\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/01\/rd-sharma-solutions-for-class-11-maths-chapter-9-14.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 9 \u2013 Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 15\"><\/p>\n<\/article>\n<p>This is the complete blog on RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.1. To Know more about the <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a>&nbsp;Class 11 Maths exam, you can ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-of-rd-sharma-solutions-class-11-maths-chapter-11\"><\/span>FAQs of RD Sharma Solutions Class 11 Maths Chapter 11<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-1630474249669\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-solutions-for-class-11-maths-chapter-9-exercise-91\"><\/span>How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are 45 questions in\u00a0RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630474338796\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-9-exercise-91\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog. <\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630474370546\" 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-for-class-11-maths-chapter-9-exercise-91\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.1: If you are looking for a good guidebook for your Class 11 Maths, then we would recommend RD Sharma Solutions Class 11 Maths. You can prepare for your exams or make your assignments with this without any hassle. To know more about the RD Sharma &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 9 Exercise 9.1 (Updated  for 2023-24)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-9-exercise-9-1\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 9 Exercise 9.1 (Updated  for 2023-24)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":122273,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,73411,73410],"tags":[3428,73334,73564],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67932"}],"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=67932"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67932\/revisions"}],"predecessor-version":[{"id":471014,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67932\/revisions\/471014"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/122273"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67932"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67932"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67932"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}