{"id":67934,"date":"2023-09-13T11:45:00","date_gmt":"2023-09-13T06:15:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67934"},"modified":"2023-11-14T10:08:34","modified_gmt":"2023-11-14T04:38:34","slug":"rd-sharma-solutions-class-11-maths-chapter-9-exercise-9-3","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-9-exercise-9-3\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 9 Exercise 9.3 (Updated For 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-122327\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-9-Exercise-9.3.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.3\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-9-Exercise-9.3.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-9-Exercise-9.3-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.3:<\/strong> Featuring a large number of worked examples and examples, <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> provides step-by-step explanations of many difficult concepts and includes a wide variety of questions to practice. Prepare for your Class 11 Maths exams by downloading 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\">Free PDF of RD Sharma Solutions Class 11 Maths Chapter 9<\/a> Exercise 9.3.<\/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-69f41a61caff3\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-9-exercise-9-3\/#download-rd-sharma-solutions-class-11-maths-chapter-9-exercise-93-pdf\" title=\"Download RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.3 PDF\">Download RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.3 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-9-3\/#access-answers-of-rd-sharma-solutions-for-class-11-maths-chapter-9-exercise-93\" title=\"Access answers of RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.3\">Access answers of RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.3<\/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-9-3\/#faqs-on-rd-sharma-solutions-for-class-11-maths-chapter-9-exercise-93\" title=\"FAQs on RD Sharma Solutions For Class 11 Maths Chapter 9 Exercise 9.3\">FAQs on RD Sharma Solutions For Class 11 Maths Chapter 9 Exercise 9.3<\/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-9-3\/#how-many-questions-are-there-in-rd-sharma-solutions-for-class-11-maths-chapter-9-exercise-93\" title=\"How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.3?\">How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.3?<\/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-9-3\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-9-exercise-93\" title=\"From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.3?\">From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.3?<\/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-9-3\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-11-maths-chapter-9-exercise-93\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.3?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.3?<\/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-93-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.3 PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-9-Ex-9.3-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 11 Solutions Chapter 9 Exercise 9.3<\/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-9.3-1.pdf\",\"#example1\");<\/script><\/p>\n<h2 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-for-class-11-maths-chapter-9-exercise-93\"><\/span>Access answers of RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.3<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Prove that:<\/strong><\/p>\n<p><strong>1. sin<sup>2<\/sup>&nbsp;2\u03c0\/5 \u2013 sin<sup>2<\/sup>&nbsp;\u03c0\/3 = (\u221a5 \u2013 1)\/8<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>sin<sup>2<\/sup>&nbsp;2\u03c0\/5 \u2013 sin<sup>2<\/sup>&nbsp;\u03c0\/3 = sin<sup>2<\/sup>&nbsp;(\u03c0\/2 \u2013 \u03c0\/10) \u2013 sin<sup>2<\/sup>&nbsp;\u03c0\/3<\/p>\n<p>we know, sin (90\u00b0\u2013 A) = cos A<\/p>\n<p>So, sin<sup>2<\/sup>&nbsp;(\u03c0\/2 \u2013 \u03c0\/10) = cos<sup>2<\/sup>&nbsp;\u03c0\/10<\/p>\n<p>Sin \u03c0\/3 = \u221a3\/2<\/p>\n<p>Then the above equation becomes,<\/p>\n<p>= Cos<sup>2<\/sup>&nbsp;\u03c0\/10 \u2013 (\u221a3\/2)<sup>2<\/sup><\/p>\n<p>We know, cos \u03c0\/10 = \u221a(10+2\u221a5)\/4<\/p>\n<p>the above equation becomes,<\/p>\n<p>= [\u221a(10+2\u221a5)\/4]<sup>2<\/sup>&nbsp;\u2013 3\/4<\/p>\n<p>= [10 + 2\u221a5]\/16 \u2013 3\/4<\/p>\n<p>= [10 + 2\u221a5 \u2013 12]\/16<\/p>\n<p>= [2\u221a5 \u2013 2]\/16<\/p>\n<p>= [\u221a5 \u2013 1]\/8<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>2. sin<sup>2<\/sup>&nbsp;24<sup>o<\/sup>&nbsp;\u2013 sin<sup>2<\/sup>&nbsp;6<sup>o<\/sup>&nbsp;= (\u221a5 \u2013 1)\/8<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>sin<sup>2<\/sup>&nbsp;24<sup>o<\/sup>&nbsp;\u2013 sin<sup>2<\/sup>&nbsp;6<sup>o<\/sup><\/p>\n<p>we know, sin (A + B) sin (A \u2013 B) = sin<sup>2<\/sup>A \u2013 sin<sup>2<\/sup>B<\/p>\n<p>Then the above equation becomes,<\/p>\n<p>sin<sup>2<\/sup>&nbsp;24<sup>o<\/sup>&nbsp;\u2013 sin<sup>2<\/sup>&nbsp;6<sup>o<\/sup>&nbsp;= sin (24<sup>o<\/sup>&nbsp;+ 6<sup>o<\/sup>) \u2013 sin (24<sup>o<\/sup>&nbsp;\u2013 6<sup>o<\/sup>)<\/p>\n<p>= sin 30<sup>o<\/sup>&nbsp;\u2013 sin 18<sup>o<\/sup><\/p>\n<p>= sin 30<sup>o<\/sup>&nbsp;\u2013 (\u221a5 \u2013 1)\/4 [since sin 18<sup>o<\/sup>&nbsp;= (\u221a5 \u2013 1)\/4]<\/p>\n<p>= 1\/2 \u00d7 (\u221a5 \u2013 1)\/4<\/p>\n<p>= (\u221a5 \u2013 1)\/8<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>3. sin<sup>2<\/sup>&nbsp;42<sup>o<\/sup>&nbsp;\u2013 cos<sup>2<\/sup>&nbsp;78<sup>o<\/sup>&nbsp;= (\u221a5 + 1)\/8<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>sin<sup>2<\/sup>&nbsp;42<sup>o<\/sup>&nbsp;\u2013 cos<sup>2<\/sup>&nbsp;78<sup>o<\/sup>&nbsp;= sin<sup>2<\/sup>&nbsp;(90<sup>o<\/sup>&nbsp;\u2013 48<sup>o<\/sup>) \u2013 cos<sup>2<\/sup>&nbsp;(90<sup>o<\/sup>&nbsp;\u2013 12<sup>o<\/sup>)<\/p>\n<p>= cos<sup>2<\/sup>&nbsp;48<sup>o<\/sup>&nbsp;\u2013 sin<sup>2<\/sup>&nbsp;12<sup>o<\/sup>&nbsp;[since, sin (90 \u2013 A) = cos A and cos (90 \u2013 A) = sin A]<\/p>\n<p>We know, cos (A + B) cos (A \u2013 B) = cos<sup>2<\/sup>A \u2013 sin<sup>2<\/sup>B<\/p>\n<p>Then the above equation becomes,<\/p>\n<p>= cos<sup>2<\/sup>&nbsp;(48<sup>o<\/sup>&nbsp;+ 12<sup>o<\/sup>) cos (48<sup>o<\/sup>&nbsp;\u2013 12<sup>o<\/sup>)<\/p>\n<p>= cos 60<sup>o<\/sup>&nbsp;cos 36<sup>o<\/sup>&nbsp;[since, cos 36<sup>o<\/sup>&nbsp;= (\u221a5 + 1)\/4]<\/p>\n<p>= 1\/2 \u00d7 (\u221a5 + 1)\/4<\/p>\n<p>= (\u221a5 + 1)\/8<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>4. cos 78<sup>o<\/sup>&nbsp;cos 42<sup>o<\/sup>&nbsp;cos 36<sup>o<\/sup>&nbsp;= 1\/8<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>cos 78<sup>o<\/sup>&nbsp;cos 42<sup>o<\/sup>&nbsp;cos 36<sup>o<\/sup><\/p>\n<p>Let us multiply and divide by 2 we get,<\/p>\n<p>cos 78<sup>o<\/sup>&nbsp;cos 42<sup>o<\/sup>&nbsp;cos 36<sup>o<\/sup>&nbsp;= 1\/2 (2 cos 78<sup>o<\/sup>&nbsp;cos 42<sup>o<\/sup>&nbsp;cos 36<sup>o<\/sup>)<\/p>\n<p>We know, 2 cos A cos B = cos (A + B) + cos (A \u2013 B)<\/p>\n<p>Then the above equation becomes,<\/p>\n<p>= 1\/2 (cos (78<sup>o<\/sup>&nbsp;+ 42<sup>o<\/sup>) + cos (78<sup>o<\/sup>&nbsp;\u2013 42<sup>o<\/sup>)) \u00d7 cos 36<sup>o<\/sup><\/p>\n<p>= 1\/2 (cos 120<sup>o<\/sup>&nbsp;+ cos 36<sup>o<\/sup>) \u00d7 cos 36<sup>o<\/sup><\/p>\n<p>= 1\/2 (cos (180<sup>o<\/sup>&nbsp;\u2013 60<sup>o<\/sup>) + cos 36<sup>o<\/sup>) \u00d7 cos 36<sup>o<\/sup><\/p>\n<p>= 1\/2 (-cos (60<sup>o<\/sup>) + cos 36<sup>o<\/sup>) \u00d7 cos 36<sup>o<\/sup>&nbsp;[since, cos(180\u00b0 \u2013 A) = \u2013 A]<\/p>\n<p>= 1\/2 (-1\/2 + (\u221a5 + 1)\/4) ((\u221a5 + 1)\/4) [since, cos 36<sup>o<\/sup>&nbsp;= (\u221a5 + 1)\/4]<\/p>\n<p>= 1\/2 (\u221a5 + 1 \u2013 2)\/4 ((\u221a5 + 1)\/4)<\/p>\n<p>= 1\/2 (\u221a5 \u2013 1)\/4) ((\u221a5 + 1)\/4)<\/p>\n<p>= 1\/2 ((\u221a5)<sup>2<\/sup>&nbsp;\u2013 1<sup>2<\/sup>)\/16<\/p>\n<p>= 1\/2 (5-1)\/16<\/p>\n<p>= 1\/2 (4\/16)<\/p>\n<p>= 1\/8<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p><strong>5. cos \u03c0\/15 cos 2\u03c0\/15 cos 4\u03c0\/15 cos 7\u03c0\/15 = 1\/16<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider LHS:<\/p>\n<p>cos \u03c0\/15 cos 2\u03c0\/15 cos 4\u03c0\/15 cos 7\u03c0\/15<\/p>\n<p>Let us multiply and divide by 2 sin \u03c0\/15, and we get,<\/p>\n<p>= [2 sin \u03c0\/15 cos \u03c0\/15] cos 2\u03c0\/15 cos 4\u03c0\/15 cos 7\u03c0\/15] \/ 2 sin \u03c0\/15<\/p>\n<p>We know, 2sin A cos A = sin 2A<\/p>\n<p>Then the above equation becomes,<\/p>\n<p>= [(sin 2\u03c0\/15) cos 2\u03c0\/15 cos 4\u03c0\/15 cos 7\u03c0\/15] \/ 2 sin \u03c0\/15<\/p>\n<p>Now, multiply and divide by 2 we get,<\/p>\n<p>= [(2 sin 2\u03c0\/15 cos 2\u03c0\/15) cos 4\u03c0\/15 cos 7\u03c0\/15] \/ 2 \u00d7 2 sin \u03c0\/15<\/p>\n<p>We know, 2sin A cos A = sin 2A<\/p>\n<p>Then the above equation becomes,<\/p>\n<p>= [(sin 4\u03c0\/15) cos 4\u03c0\/15 cos 7\u03c0\/15] \/ 4 sin \u03c0\/15<\/p>\n<p>Now, multiply and divide by 2, and we get,<\/p>\n<p>= [(2 sin 4\u03c0\/15 cos 4\u03c0\/15) cos 7\u03c0\/15] \/ 2 \u00d7 4 sin \u03c0\/15<\/p>\n<p>We know, 2sin A cos A = sin 2A<\/p>\n<p>Then the above equation becomes,<\/p>\n<p>= [(sin 8\u03c0\/15) cos 7\u03c0\/15] \/ 8 sin \u03c0\/15<\/p>\n<p>Now, multiply and divide by 2, and we get,<\/p>\n<p>= [2 sin 8\u03c0\/15 cos 7\u03c0\/15] \/ 2 \u00d7 8 sin \u03c0\/15<\/p>\n<p>We know, 2sin A cos B = sin (A+B) + sin (A\u2013B)<\/p>\n<p>Then the above equation becomes,<\/p>\n<p>= [sin (8\u03c0\/15 + 7\u03c0\/15) + sin (8\u03c0\/15 \u2013 7\u03c0\/15)] \/ 16 sin \u03c0\/15<\/p>\n<p>= [sin (\u03c0) + sin (\u03c0\/15)] \/ 16 sin \u03c0\/15<\/p>\n<p>= [0 + sin (\u03c0\/15)] \/ 16 sin \u03c0\/15<\/p>\n<p>= sin (\u03c0\/15) \/ 16 sin \u03c0\/15<\/p>\n<p>= 1\/16<\/p>\n<p>= RHS<\/p>\n<p>Hence proved.<\/p>\n<p>This is the complete blog of RD Sharma Solutions For Class 11 Maths Chapter 9 Exercise 9.3. 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.&nbsp;<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-for-class-11-maths-chapter-9-exercise-93\"><\/span>FAQs on RD Sharma Solutions For Class 11 Maths Chapter 9 Exercise 9.3<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630476768085\" 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-93\"><\/span>How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.3?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are 10 questions in\u00a0RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.3.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630476834971\" 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-93\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.3?<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-1630476859269\" 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-93\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 9 Exercise 9.3?<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 9.3: Featuring a large number of worked examples and examples, RD Sharma Solutions Class 11 Maths provides step-by-step explanations of many difficult concepts and includes a wide variety of questions to practice. Prepare for your Class 11 Maths exams by downloading the Free PDF of RD &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 9 Exercise 9.3 (Updated For 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-9-exercise-9-3\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 9 Exercise 9.3 (Updated For 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":122327,"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\/67934"}],"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=67934"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67934\/revisions"}],"predecessor-version":[{"id":506793,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67934\/revisions\/506793"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/122327"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67934"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67934"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67934"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}