{"id":67925,"date":"2023-09-07T22:55:00","date_gmt":"2023-09-07T17:25:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67925"},"modified":"2023-12-12T12:15:19","modified_gmt":"2023-12-12T06:45:19","slug":"rd-sharma-solutions-class-11-maths-chapter-7-exercise-7-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-7-exercise-7-2\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 7 Exercise 7.2 (Updated For 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-122199\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-7-Exercise-7.2.jpg\" alt=\"RD Sharma Class 11 Solutions Chapter 7 Exercise 7.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-7-Exercise-7.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-7-Exercise-7.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 11 Solutions Chapter 7 Exercise 7.2: <\/strong>To ace your Class 11 Maths exam you must study from <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 download the Free PDF of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-7-trigonometric-ratios-of-compound-angles\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 7<\/a> Exercise 7.2. All your doubts will be cleared easily here.<\/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-69e233075c4ee\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-7-exercise-7-2\/#how-many-questions-are-there-in-rd-sharma-solutions-for-class-11-maths-chapter-7-exercise-72\" title=\"How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 7 Exercise 7.2?\">How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 7 Exercise 7.2?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-7-exercise-72-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 7 Exercise 7.2 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-7-Ex-7.2-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 11 Solutions Chapter 7 Exercise 7.2<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-7-Ex-7.2-1.pdf\",\"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-11-maths-chapter-7-exercise-72\"><\/span>RD Sharma Solutions Class 11 Maths Chapter 7 Exercise 7.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>1. Find the maximum and minimum values of each of the following trigonometrical expressions:<\/strong><\/p>\n<p><strong>(i) 12 sin x \u2013 5 cos x<br \/>(ii) 12 cos x + 5 sin x + 4<br \/>(iii) 5 cos x + 3 sin (\u03c0\/6 \u2013 x) + 4<br \/>(iv) sin x \u2013 cos x + 1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that the maximum value of A cos \u03b1 + B sin \u03b1 + C is C + \u221a(A<sup>2<\/sup>\u00a0+B<sup>2<\/sup>),<\/p>\n<p>And the minimum value is C \u2013 \u221a(a<sup>2<\/sup>\u00a0+B<sup>2<\/sup>).<\/p>\n<p><strong>(i)<\/strong>\u00a012 sin x \u2013 5 cos x<\/p>\n<p>Given: f(x) = 12 sin x \u2013 5 cos x<\/p>\n<p>Here, A = -5, B = 12 and C = 0<\/p>\n<p>\u2013<strong>\u221a<\/strong>((-5)<sup>2<\/sup>\u00a0+ 12<sup>2<\/sup>) \u2264 12 sin x \u2013 5 cos x \u2264\u00a0<strong>\u221a<\/strong>((-5)<sup>2<\/sup>\u00a0+ 12<sup>2<\/sup>)<\/p>\n<p>\u2013<strong>\u221a<\/strong>(25+144) \u2264 12 sin x \u2013 5 cos x \u2264\u00a0<strong>\u221a<\/strong>(25+144)<\/p>\n<p>\u2013<strong>\u221a<\/strong>169 \u2264 12 sin x \u2013 5 cos x \u2264\u00a0<strong>\u221a<\/strong>169<\/p>\n<p>-13\u00a0\u2264\u00a012 sin\u00a0x \u2013 5 cos\u00a0x \u2264 13<\/p>\n<p>Hence, the maximum and minimum values of f(x) are 13 and -13, respectively.<\/p>\n<p><strong>(ii)\u00a0<\/strong>12 cos x + 5 sin x + 4<\/p>\n<p>Given: f(x) = 12 cos x + 5 sin x + 4<\/p>\n<p>Here, A = 12, B = 5 and C = 4<\/p>\n<p>4 \u2013\u00a0<strong>\u221a<\/strong>(12<sup>2<\/sup>\u00a0+ 5<sup>2<\/sup>) \u2264 12 cos x + 5 sin x + 4 \u2264 4 +\u00a0<strong>\u221a<\/strong>(12<sup>2<\/sup>\u00a0+ 5<sup>2<\/sup>)<\/p>\n<p>4 \u2013\u00a0<strong>\u221a<\/strong>(144+25) \u2264 12 cos x + 5 sin x + 4 \u2264 4 +\u00a0<strong>\u221a<\/strong>(144+25)<\/p>\n<p>4 \u2013<strong>\u221a<\/strong>169 \u2264 12 cos x + 5 sin x + 4 \u2264 4 +\u00a0<strong>\u221a<\/strong>169<\/p>\n<p>-9\u00a0\u2264\u00a012 cos x + 5 sin x + 4 \u2264 17<\/p>\n<p>Hence, the maximum and minimum values of f(x) are -9 and 17, respectively.<\/p>\n<p><strong>(iii)\u00a0<\/strong>5 cos x + 3 sin (\u03c0\/6 \u2013 x) + 4<strong>\u00a0<\/strong><\/p>\n<p>Given: f(x) = 5 cos x + 3 sin (\u03c0\/6 \u2013 x) + 4<strong>\u00a0<\/strong><\/p>\n<p>We know that sin (A \u2013 B) = sin A cos B \u2013 cos A sin B<\/p>\n<p>f(x) = 5 cos x + 3 sin (\u03c0\/6 \u2013 x) + 4<strong>\u00a0<\/strong><\/p>\n<p><strong>=<\/strong>\u00a05 cos x + 3 (sin \u03c0\/6 cos x \u2013 cos \u03c0\/6 sin x) + 4<\/p>\n<p>= 5 cos x + 3\/2 cos x \u2013 3<strong>\u221a<\/strong>3\/2 sin x + 4<\/p>\n<p>= 13\/2 cos x \u2013 3<strong>\u221a<\/strong>3\/2 sin x + 4<\/p>\n<p>So, here A = 13\/2, B = \u2013 3<strong>\u221a<\/strong>3\/2, C = 4<\/p>\n<p>4 \u2013\u00a0<strong>\u221a<\/strong>[(13\/2)<sup>2<\/sup>\u00a0+ (-3<strong>\u221a<\/strong>3\/2)<sup>2<\/sup>] \u2264 13\/2 cos x \u2013 3<strong>\u221a<\/strong>3\/2 sin x + 4 \u2264 4 +\u00a0<strong>\u221a<\/strong>[(13\/2)<sup>2<\/sup>\u00a0+ (-3<strong>\u221a<\/strong>3\/2)<sup>2<\/sup>]<\/p>\n<p>4 \u2013\u00a0<strong>\u221a<\/strong>[(169\/4) + (27\/4)] \u2264 13\/2 cos x \u2013 3<strong>\u221a<\/strong>3\/2 sin x + 4 \u2264 4 +\u00a0<strong>\u221a<\/strong>[(169\/4) + (27\/4)]<\/p>\n<p>4 \u2013 7 \u2264 13\/2 cos x \u2013 3<strong>\u221a<\/strong>3\/2 sin x + 4 \u2264 4 + 7<\/p>\n<p>-3 \u2264 13\/2 cos x \u2013 3<strong>\u221a<\/strong>3\/2 sin x + 4 \u2264 11<\/p>\n<p>Hence, the maximum and minimum values of f(x) are -3 and 11, respectively.<\/p>\n<p><strong>(iv)<\/strong>\u00a0sin x \u2013 cos x + 1<\/p>\n<p>Given: f(x) = sin x \u2013 cos x + 1<\/p>\n<p>So, here A = -1, B = 1 And c = 1<\/p>\n<p>1 \u2013\u00a0<strong>\u221a<\/strong>[(-1)<sup>2<\/sup>\u00a0+ 1<sup>2<\/sup>] \u2264 sin x \u2013 cos x + 1 \u2264 1 +\u00a0<strong>\u221a<\/strong>[(-1)<sup>2<\/sup>\u00a0+ 1<sup>2<\/sup>]<\/p>\n<p>1 \u2013\u00a0<strong>\u221a<\/strong>(1+1) \u2264 sin x \u2013 cos x + 1 \u2264 1 +\u00a0<strong>\u221a<\/strong>(1+1)<\/p>\n<p>1 \u2013\u00a0<strong>\u221a<\/strong>2 \u2264 sin x \u2013 cos x + 1 \u2264 1 +\u00a0<strong>\u221a<\/strong>2<\/p>\n<p>Hence, the maximum and minimum values of f(x) are 1 \u2013\u00a0<strong>\u221a<\/strong>2 and 1 +\u00a0<strong>\u221a<\/strong>2, respectively.<\/p>\n<p><strong>2. Reduce each of the following expressions to the Sine and Cosine of a single expression:<\/strong><\/p>\n<p><strong>(i) \u221a3 sin x \u2013 cos x<\/strong><\/p>\n<p><strong>(ii) cos x \u2013 sin x<\/strong><\/p>\n<p><strong>(iii) 24 cos x + 7 sin x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>\u221a3 sin x \u2013 cos x<\/p>\n<p>Let f(x) = \u221a3 sin x \u2013 cos x<\/p>\n<p>Dividing and multiplying by \u221a((\u221a3)<sup>2<\/sup>\u00a0+ 1<sup>2<\/sup>) i.e. by 2<\/p>\n<p>f(x) = 2(\u221a3\/2 sin x \u2013 1\/2 cos x)<\/p>\n<p>Sine expression:<\/p>\n<p>f(x) = 2(cos \u03c0\/6 sin x \u2013 sin \u03c0\/6 cos x) (since, \u221a3\/2 = cos \u03c0\/6 and 1\/2 = sin \u03c0\/6)<\/p>\n<p>We know that sin A cos B \u2013 cos A sin B = sin (A \u2013 B)<\/p>\n<p>f(x) = 2 sin (x \u2013 \u03c0\/6)<\/p>\n<p>Again,<\/p>\n<p>f(x) = 2(\u221a3\/2 sin x \u2013 1\/2 cos x)<\/p>\n<p>Cosine expression:<\/p>\n<p>f(x) = 2(sin \u03c0\/3 sin x \u2013 cos \u03c0\/3 cos x)<\/p>\n<p>We know that cos A cos B \u2013 sin A sin B = cos (A + B)<\/p>\n<p>f(x) = -2 cos(\u03c0\/3 + x)<\/p>\n<p><strong>(ii)\u00a0<\/strong>cos x \u2013 sin x<\/p>\n<p>Let f(x) = cos x \u2013 sin x<\/p>\n<p>Dividing and multiplying by \u221a(1<sup>2<\/sup>\u00a0+ 1<sup>2<\/sup>) i.e. by \u221a2,<\/p>\n<p>f(x) = \u221a2(1\/\u221a2 cos x \u2013 1\/\u221a2 sin x)<\/p>\n<p>Sine expression:<\/p>\n<p>f(x) = \u221a2(sin \u03c0\/4 cos x \u2013 cos \u03c0\/4 sin x) (since, 1\/\u221a2 = sin \u03c0\/4 and 1\/\u221a2 = cos \u03c0\/4)<\/p>\n<p>We know that sin A cos B \u2013 cos A sin B = sin (A \u2013 B)<\/p>\n<p>f(x) = \u221a2 sin (\u03c0\/4 \u2013 x)<\/p>\n<p>Again,<\/p>\n<p>f(x) = \u221a2(1\/\u221a2 cos x \u2013 1\/\u221a2 sin x)<\/p>\n<p>Cosine expression:<\/p>\n<p>f(x) = 2(cos \u03c0\/4 cos x \u2013 sin \u03c0\/4 sin x)<\/p>\n<p>We know that cos A cos B \u2013 sin A sin B = cos (A + B)<\/p>\n<p>f(x) = \u221a2 cos (\u03c0\/4 + x)<\/p>\n<p><strong>(iii)\u00a0<\/strong>24 cos x + 7 sin x<\/p>\n<p>Let f(x) = 24 cos x + 7 sin x<\/p>\n<p>Dividing and multiplying by \u221a((\u221a24)<sup>2<\/sup>\u00a0+ 7<sup>2<\/sup>) = \u221a625 i.e. by 25,<\/p>\n<p>f(x) = 25(24\/25 cos x + 7\/25 sin x)<\/p>\n<p>Sine expression:<\/p>\n<p>f(x) = 25(sin \u03b1 cos x + cos \u03b1 sin x)\u00a0where, sin \u03b1 = 24\/25 and cos \u03b1 = 7\/25<\/p>\n<p>We know that sin A cos B + cos A sin B = sin (A + B)<\/p>\n<p>f(x) = 25 sin (\u03b1 + x)<\/p>\n<p>Cosine expression:<\/p>\n<p>f(x) = 25(cos \u03b1 cos x + sin \u03b1 sin x)\u00a0where, cos \u03b1 = 24\/25 and sin \u03b1 = 7\/25<\/p>\n<p>We know that cos A cos B + sin A sin B = cos (A \u2013 B)<\/p>\n<p>f(x) = 25 cos (\u03b1 \u2013 x)<\/p>\n<p><strong>3. Show that Sin 100<sup>o<\/sup>\u00a0\u2013 Sin 10<sup>o<\/sup>\u00a0is positive.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let f(x) = sin 100\u00b0 \u2013 sin 10\u00b0<\/p>\n<p>Dividing And multiplying by \u221a(1<sup>2<\/sup>\u00a0+ 1<sup>2<\/sup>) i.e. by \u221a2,<\/p>\n<p>f(x) = \u221a2(1\/\u221a2 sin 100<sup>o<\/sup>\u00a0\u2013 1\/\u221a2 sin 10<sup>o<\/sup>)<\/p>\n<p>f(x) = \u221a2(cos \u03c0\/4 sin (90+10)<sup>o<\/sup>\u00a0\u2013 sin \u03c0\/4 sin 10<sup>o<\/sup>) (since, 1\/\u221a2 = cos \u03c0\/4 and 1\/\u221a2 = sin \u03c0\/4)<\/p>\n<p>f(x) = \u221a2(cos \u03c0\/4 cos 10<sup>o<\/sup>\u00a0\u2013 sin \u03c0\/4 sin 10<sup>o<\/sup>)<\/p>\n<p>We know that cos A cos B \u2013 sin A sin B = cos (A + B)<\/p>\n<p>f(x) = \u221a2 cos (\u03c0\/4 + 10<sup>o<\/sup>)<\/p>\n<p>\u2234\u00a0f(x) = \u221a2 cos 55\u00b0<\/p>\n<p><strong>4. Prove that (2\u221a3 + 3) sin x + 2\u221a3 cos x\u00a0lies between \u2013 (2\u221a3 + \u221a15) and (2\u221a3 + \u221a15).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let f(x) = (2\u221a3 + 3) sin x + 2\u221a3 cos x<\/p>\n<p>Here, A = 2\u221a3, B = 2\u221a3 + 3 and C = 0<\/p>\n<p>\u2013 \u221a[(2\u221a3)<sup>2<\/sup>\u00a0+ (2\u221a3 + 3)<sup>2<\/sup>] \u2264 (2\u221a3 + 3) sin x + 2\u221a3 cos x \u2264 \u221a[(2\u221a3)<sup>2<\/sup>\u00a0+ (2\u221a3 + 3)<sup>2<\/sup>]<\/p>\n<p>\u2013 \u221a[12+12+9+12\u221a3] \u2264 (2\u221a3 + 3) sin x + 2\u221a3 cos x \u2264 \u221a[12+12+9+12\u221a3]<\/p>\n<p>\u2013 \u221a[33+12\u221a3] \u2264 (2\u221a3 + 3) sin x + 2\u221a3 cos x \u2264 \u221a[33+12\u221a3]<\/p>\n<p>\u2013 \u221a[15+12+6+12\u221a3] \u2264 (2\u221a3 + 3) sin x + 2\u221a3 cos x \u2264 \u221a[15+12+6+12\u221a3]<\/p>\n<p>We know that (12\u221a3 + 6 &lt; 12\u221a5) because the value of \u221a5 \u2013 \u221a3 is more than 0.5<\/p>\n<p>So if we replace (12\u221a3 + 6 with 12\u221a5), the above inequality still holds.<\/p>\n<p>So by rearranging the above expression \u221a(15+12+12\u221a5)we get, 2\u221a3 + \u221a15<\/p>\n<p>\u2013 2\u221a3 + \u221a15 \u2264 (2\u221a3 + 3) sin x + 2\u221a3 cos x \u2264 2\u221a3 + \u221a15<\/p>\n<p>Hence proved.<\/p>\n<p>This is the complete blog on RD Sharma Solutions For Class 11 Maths Chapter 7 Exercise 7.2. To Know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a>. Class 11 Maths exam, ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-for-class-11-maths-chapter-7-exercise-72\"><\/span>FAQs on RD Sharma Solutions For Class 11 Maths Chapter 7 Exercise 7.2<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-1630430548730\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-even-beneficial-to-study-rd-sharma-solutions-class-11-maths-chapter-7-exercise-72\"><\/span>Is it even beneficial to study RD Sharma Solutions Class 11 Maths Chapter 7 Exercise 7.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, your preparation will be strengthened with this amazing help book. All your questions will be answered by this book.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630430599923\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-for-class-11-maths-chapter-7-exercise-72-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 11 Maths Chapter 7 Exercise 7.2\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630430627589\" 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-7-exercise-72\"><\/span>How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 7 Exercise 7.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are 4 questions in\u00a0RD Sharma Solutions Class 11 Maths Chapter 7 Exercise 7.2.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 11 Solutions Chapter 7 Exercise 7.2: To ace your Class 11 Maths exam you must study from RD Sharma Solutions Class 11 Maths. You can download the Free PDF of RD Sharma Solutions Class 11 Maths Chapter 7 Exercise 7.2. All your doubts will be cleared easily here. Download RD Sharma Solutions &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 7 Exercise 7.2 (Updated For 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-7-exercise-7-2\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 7 Exercise 7.2 (Updated For 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":122204,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3428,73334,73564],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67925"}],"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=67925"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67925\/revisions"}],"predecessor-version":[{"id":520908,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67925\/revisions\/520908"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/122204"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67925"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67925"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67925"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}