{"id":6070,"date":"2023-09-08T04:36:25","date_gmt":"2023-09-08T04:36:25","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6070"},"modified":"2023-09-08T04:36:25","modified_gmt":"2023-09-08T04:36:25","slug":"ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 8 (Introduction to Trigonometry)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 8 Introduction to Trigonometry\u00a0 <\/strong>Exercise 8.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 8 Introduction to Trigonometry<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 8.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 8.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-4\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 8.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 8.2<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Evaluate the following:<br \/>\n<\/strong>(i) sin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0<br \/>\n(ii) 2 tan<sup>2<\/sup> 45\u00b0 + cos<sup>2<\/sup> 30\u00b0 \u2013 sin<sup>2<\/sup> 60<br \/>\n(iii) cos 45\u00b0\/(sec 30\u00b0 + cosec 30\u00b0)<br \/>\n(iv) (sin 30\u00b0 + tan 45\u00b0 \u2013 cosec 60\u00b0)\/(sec 30\u00b0 + cos 60\u00b0 + cot 45\u00b0)<br \/>\n(v) (5 cos<sup>2<\/sup>60\u00b0 + 4 sec<sup>2<\/sup>30\u00b0\u00a0&#8211; tan<sup>2<\/sup>45\u00b0)\/(sin<sup>2<\/sup>30\u00b0\u00a0+\u00a0cos<sup>2<\/sup>30\u00b0)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<\/strong> \u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) sin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0<br \/>\n<\/strong>Using the values of the given trigonometric ratios<br \/>\nsin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0<br \/>\n= \u221a3\/2 \u00d7 \u221a3\/2 + (1\/2) \u00d7 (1\/2 )<br \/>\n= 3\/4 + 1\/4<br \/>\n= 4\/4<br \/>\n=1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 2 tan<sup>2<\/sup> 45\u00b0 + cos<sup>2<\/sup> 30\u00b0 \u2013 sin<sup>2<\/sup> 60<br \/>\n<\/strong>Using the values of the given trigonometric ratios<br \/>\n2 tan<sup>2<\/sup> 45\u00b0 + cos<sup>2<\/sup> 30\u00b0 \u2013 sin<sup>2<\/sup> 60<br \/>\n= 2(1)<sup>2 <\/sup>+ (\u221a3\/2)<sup>2 <\/sup>&#8211; (\u221a3\/2)<sup>2<br \/>\n<\/sup>= 2 + 0<br \/>\n= 2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) cos 45\u00b0\/ (sec 30\u00b0+cosec 30\u00b0)<br \/>\n<\/strong>Using the values of the given trigonometric ratios<br \/>\n= <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{1}{\\sqrt{2}}}{\\frac{2}{\\sqrt{3}}+2}\" alt=\"\\frac{\\frac{1}{\\sqrt{2}}}{\\frac{2}{\\sqrt{3}}+2}\" width=\"52\" height=\"55\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{1}{\\sqrt{2}}}{\\frac{2&amp;space;+2\\sqrt{3}}{\\sqrt{3}}}\" alt=\"\\frac{\\frac{1}{\\sqrt{2}}}{\\frac{2 +2\\sqrt{3}}{\\sqrt{3}}}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{\\sqrt{2}}&amp;space;\\times&amp;space;\\frac{\\sqrt{3}}{2&amp;space;+2\\sqrt{3}}\" alt=\"\\frac{1}{\\sqrt{2}} \\times \\frac{\\sqrt{3}}{2 +2\\sqrt{3}}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\sqrt{3}}{2\\sqrt{2}(\\sqrt{3}+1)}\" alt=\"\\frac{\\sqrt{3}}{2\\sqrt{2}(\\sqrt{3}+1)}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Now, rationalize the terms we get\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\sqrt{3}}{2\\sqrt{2}(\\sqrt{3}+1)}\" alt=\"\\frac{\\sqrt{3}}{2\\sqrt{2}(\\sqrt{3}+1)}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\sqrt{3}&amp;space;-1}{\\sqrt{3}&amp;space;-1}\" alt=\"\\frac{\\sqrt{3} -1}{\\sqrt{3} -1}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3-\\sqrt{3}}{4\\sqrt{2}}\" alt=\"\\frac{3-\\sqrt{3}}{4\\sqrt{2}}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Now, multiply both the numerator and denominator by \u221a2 , we get\u00a0 <\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3-\\sqrt{3}}{4\\sqrt{2}}\" alt=\"\\frac{3-\\sqrt{3}}{4\\sqrt{2}}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\sqrt{2}}{\\sqrt{2}}\" alt=\"\\frac{\\sqrt{2}}{\\sqrt{2}}\" align=\"absmiddle\" \/>\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3\\sqrt{2}-\\sqrt{6}}{8}\" alt=\"\\frac{3\\sqrt{2}-\\sqrt{6}}{8}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iv) (sin 30\u00b0 + tan 45\u00b0 \u2013 cosec 60\u00b0)\/(sec 30\u00b0 + cos 60\u00b0 + cot 45\u00b0)<br \/>\n<\/strong>Using the values of the given trigonometric ratios <\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{1}{2}&amp;space;+&amp;space;1&amp;space;-\\frac{2}{\\sqrt{3}}}{\\frac{2}{\\sqrt{3}}+\\frac{1}{2}&amp;space;+1}\" alt=\"\\frac{\\frac{1}{2} + 1 -\\frac{2}{\\sqrt{3}}}{\\frac{2}{\\sqrt{3}}+\\frac{1}{2} +1}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{\\sqrt{3}+2\\sqrt{3}-4}{2\\sqrt{3}}}{\\frac{4+\\sqrt{3+2\\sqrt{3}}}{2\\sqrt{3}}}\" alt=\"\\frac{\\frac{\\sqrt{3}+2\\sqrt{3}-4}{2\\sqrt{3}}}{\\frac{4+\\sqrt{3+2\\sqrt{3}}}{2\\sqrt{3}}}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3\\sqrt{3}-4}{3\\sqrt{3}+4}\" alt=\"\\frac{3\\sqrt{3}-4}{3\\sqrt{3}+4}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p><span style=\"color: #000000;\">Now, rationalize the terms we get\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3\\sqrt{3}-4}{3\\sqrt{3}+4}\" alt=\"\\frac{3\\sqrt{3}-4}{3\\sqrt{3}+4}\" width=\"65\" height=\"46\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3\\sqrt{3}-4}{3\\sqrt{3}-4}\" alt=\"\\frac{3\\sqrt{3}-4}{3\\sqrt{3}-4}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{27-12\\sqrt{3}-12\\sqrt{3}+16}{27-12\\sqrt{3}&amp;space;+12\\sqrt{3}+16}\" alt=\"\\frac{27-12\\sqrt{3}-12\\sqrt{3}+16}{27-12\\sqrt{3} +12\\sqrt{3}+16}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{27-14\\sqrt{3}+16}{11}\" alt=\"\\frac{27-14\\sqrt{3}+16}{11}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{43-24\\sqrt{3}}{11}\" alt=\"\\frac{43-24\\sqrt{3}}{11}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(v) (5cos<sup>2<\/sup>60\u00b0\u00a0+\u00a04sec<sup>2<\/sup>30\u00b0\u00a0&#8211; tan<sup>2<\/sup>45\u00b0)\/(sin<sup>2<\/sup>30\u00b0\u00a0+\u00a0cos<sup>2<\/sup>30\u00b0)<br \/>\n<\/strong>Using the values of the given trigonometric ratios<br \/>\n= 5(1\/2)<sup>2 <\/sup>+ 4(2\/\u221a3)<sup>2 <\/sup>&#8211; 1<sup>2<\/sup>\/(1\/2)<sup>2 <\/sup>+ (\u221a3\/2)<sup>2<\/sup><sup><br \/>\n<\/sup>= (5\/4 + 16\/3 &#8211; 1)\/(1\/4 + 3\/4)<br \/>\n= (15 + 64 &#8211; 12)\/12\/(4\/4)<br \/>\n= 67\/12<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Choose the correct option and justify your choice :<br \/>\n<\/strong><strong>(i) 2tan 30\u00b0\/1+tan<sup>2<\/sup>30\u00b0 =<br \/>\n<\/strong>(A) sin 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(B) cos 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C) tan 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(D) sin 30\u00b0<br \/>\n<strong>(ii) 1-tan<sup>2<\/sup>45\u00b0\/1+tan<sup>2<\/sup>45\u00b0 =<br \/>\n<\/strong>(A) tan 90\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(B) 1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C) sin 45\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(D) 0<br \/>\n<strong>(iii)\u00a0\u00a0sin 2A = 2 sin A is true when A =<br \/>\n<\/strong>(A) 0\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (B) 30\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C) 45\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (D) 60\u00b0<br \/>\n<strong>(iv) 2tan30\u00b0\/1-tan<sup>2<\/sup>30\u00b0 =<br \/>\n<\/strong>(A) cos 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(B) sin 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (C) tan 60\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (D) sin 30\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<\/strong>\u00a0 \u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) 2tan 30\u00b0\/1+tan<sup>2<\/sup>30\u00b0 =<br \/>\n<\/strong>= 2(1\/\u221a3)\/1 + (1\/\u221a3)<sup>2<br \/>\n<\/sup>= (2\/\u221a3)\/(1 + 1\/3)<br \/>\n= (2\/\u221a3)\/(4\/3)<br \/>\n= 6\/4\u221a3<br \/>\n= \u221a3\/2<br \/>\n= sin 60\u00b0<br \/>\n<strong>(A) is correct.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 1 &#8211; tan<sup>2<\/sup>45\u00b0\/1 + tan<sup>2<\/sup>45\u00b0 =<br \/>\n<\/strong>= (1 &#8211; 1<sup>2<\/sup>)\/(1 + 1<sup>2<\/sup>)<br \/>\n= 0\/2<br \/>\n= 0<br \/>\n<strong>(D) is correct.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii)\u00a0\u00a0sin 2A = 2 sin A is true when A =<br \/>\n<\/strong>sin 2A = sin 0\u00b0 = 0<br \/>\n2 sin A = 2 sin 0\u00b0<br \/>\n= 2 \u00d7 0<br \/>\n= 0<br \/>\nsin 2A = 2sin A cos A<br \/>\n\u21d22sin A cos A =\u00a02 sin A<br \/>\n\u21d2 2cos A = 2<br \/>\n\u21d2 cos A = 1<br \/>\nTherefore,<br \/>\n\u21d2 A = 0\u00b0<br \/>\n<strong>(A) is correct.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) 2tan30\u00b0\/1-tan<sup>2<\/sup>30\u00b0 =<br \/>\n<\/strong>= \u00a02(1\/\u221a3)\/1 &#8211; (1\/\u221a3)<sup>2<br \/>\n<\/sup>= (2\/\u221a3)\/(1 &#8211; 1\/3)<br \/>\n= (2\/\u221a3)\/(2\/3)<br \/>\n= \u221a3<br \/>\n= tan 60\u00b0<br \/>\n<strong>(C) is correct.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. If tan (A + B) = \u221a3 and tan (A \u2013 B) = 1\/\u221a3 ,0\u00b0 &lt; A + B \u2264 90\u00b0; A &gt; B, find A and B.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<\/strong><br \/>\ntan (A + B) = \u221a3<br \/>\n\u221a3 = tan 60\u00b0<br \/>\nNow substitute the degree value<br \/>\ntan (A + B) = tan 60\u00b0<br \/>\n(A + B) = 60\u00b0\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8211; (i)<br \/>\nThe above equation is assumed as equation (i)<br \/>\ntan (A \u2013 B) = 1\/\u221a3<br \/>\n1\/\u221a3 = tan 30\u00b0<br \/>\nNow substitute the degree value<br \/>\ntan (A \u2013 B) = tan 30\u00b0<br \/>\n(A \u2013 B) = 30\u00b0\u00a0 \u00a0&#8212;&#8212;&#8212;&#8211; (ii)<br \/>\nNow add the equation (i) and (ii), we get<br \/>\nA + B + A \u2013 B = 60\u00b0 + 30\u00b0<br \/>\n2A = 90\u00b0<br \/>\nA = 45\u00b0<br \/>\nNow, substitute the value of A in equation (i) to find the value of B<br \/>\n45\u00b0\u00a0+ B = 60\u00b0<br \/>\nB = 60\u00b0 \u2013 45\u00b0<br \/>\nB = 15\u00b0<br \/>\nTherefore<br \/>\nA = 45\u00b0<br \/>\nB = 15\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. State whether the following are true or false. Justify your answer.<br \/>\n<\/strong>(i) sin (A + B) = sin A + sin B.<br \/>\n(ii) The value of sin \u03b8 increases as \u03b8 increases.<br \/>\n(iii) The value of cos \u03b8 increases as \u03b8 increases.<br \/>\n(iv) sin \u03b8 = cos \u03b8 for all values of \u03b8.<br \/>\n(v) cot A is not defined for A = 0\u00b0.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<\/strong> \u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) sin (A + B) = sin A + sin B.<br \/>\n<\/strong>Let A = 30\u00b0 and B = 60\u00b0, then <\/span><br \/>\n<span style=\"color: #000000;\">sin (A + B) = sin (30\u00b0 + 60\u00b0) = sin 90\u00b0 = 1\u00a0 <\/span><br \/>\n<span style=\"color: #000000;\">sin A + sin B = sin 30\u00b0 + sin 60\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">= 1\/2 + \u221a3\/2 <\/span><br \/>\n<span style=\"color: #000000;\">= (1 + \u221a3)\/2<\/span><br \/>\n<span style=\"color: #000000;\"><strong>False<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) The value of sin \u03b8 increases as \u03b8 increases.<br \/>\n<\/strong>sin 0\u00b0 = 0<\/span><br \/>\n<span style=\"color: #000000;\">sin 30\u00b0 = 1\/2<\/span><br \/>\n<span style=\"color: #000000;\">sin 45\u00b0 = 1\/\u221a2<\/span><br \/>\n<span style=\"color: #000000;\">sin\u00a060\u00b0 =\u00a0\u221a3\/2<br \/>\nsin \u00a090\u00b0 = 1 <\/span><br \/>\n<span style=\"color: #000000;\">Thus the value of sin \u03b8 increases as \u03b8 increases.<\/span><br \/>\n<span style=\"color: #000000;\"><strong>True<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">(iii) The value of cos \u03b8 increases as \u03b8 increases.<br \/>\ncos 0\u00b0 = 1<br \/>\ncos 30\u00b0 = \u221a3\/2<br \/>\ncos 45\u00b0 = 1\/\u221a2<br \/>\ncos 60\u00b0 =\u00a01\/2<br \/>\ncos 90\u00b0 = 0<br \/>\nThus, the value of cos \u03b8 decreases as \u03b8 increases. So, the statement given above is <strong>False<\/strong>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) sin \u03b8 = cos \u03b8 for all values of \u03b8.<br \/>\n<\/strong>sin \u03b8 = cos \u03b8,<br \/>\nwhen a right triangle has 2 angles of (\u03c0\/4).<br \/>\nTherefore, the above statement is <strong>False<\/strong>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) cot A is not defined for A = 0\u00b0.<br \/>\n<\/strong>cot A = cos A\/sin A<br \/>\nNow substitute A = 0\u00b0<br \/>\ncot 0\u00b0 = cos 0\u00b0\/sin 0\u00b0 = 1\/0 = undefined.<br \/>\nHence, it is <strong>True<\/strong><\/span><\/p>\n<p><span style=\"color: #0000ff; font-family: Georgia, Palatino;\"><b><i><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\">Go Back To Chapters<\/a><\/i><\/b><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 8 (Introduction to Trigonometry)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 8 Introduction to Trigonometry\u00a0 Exercise 8.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 8 Introduction to Trigonometry NCERT Class 10 Maths Solution [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1091,1093,1044,1049,1048],"class_list":["post-6070","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-8-introduction-to-trigonometry-solutions","tag-ncert-class-10-mathematics-exercise-8-2-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 10 Maths\u00a0Chapter - 8 (Introduction to Trigonometry)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 8 Introduction to Trigonometry\u00a0 Exercise 8.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 8.2\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2\" \/>\n<meta property=\"og:description\" content=\"NCERT Solutions Class 10 Maths\u00a0Chapter - 8 (Introduction to Trigonometry)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 8 Introduction to Trigonometry\u00a0 Exercise 8.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 8.2\" \/>\n<meta property=\"og:url\" content=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/\" \/>\n<meta property=\"og:site_name\" content=\"TheExamPillar NCERT\" \/>\n<meta property=\"article:published_time\" content=\"2023-09-08T04:36:25+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/latex.codecogs.com\/gif.latex?fracfrac1sqrt2frac2sqrt3+2\" \/>\n<meta name=\"author\" content=\"Admin\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@exampillar\" \/>\n<meta name=\"twitter:site\" content=\"@exampillar\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Admin\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"8 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/#article\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/\"},\"author\":{\"name\":\"Admin\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"headline\":\"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2\",\"datePublished\":\"2023-09-08T04:36:25+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/\"},\"wordCount\":594,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"image\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/latex.codecogs.com\\\/gif.latex?\\\\frac{\\\\frac{1}{\\\\sqrt{2}}}{\\\\frac{2}{\\\\sqrt{3}}+2}\",\"keywords\":[\"Class 10 NCERT Mathematics Solutions\",\"NCERT Class 10 Mathematics\u00a0Chapter 8 Introduction to Trigonometry Solutions\",\"NCERT Class 10 Mathematics Exercise 8.2 Solutions\",\"NCERT Class 10 Mathematics Solutions Class 10 NCERT Solutions\",\"NCERT Solutions Class 10 Mathematics\",\"NCERT Solutions Class 10 Maths\"],\"articleSection\":[\"Class 10 Maths\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/\",\"name\":\"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 | TheExamPillar NCERT\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/#primaryimage\"},\"image\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/latex.codecogs.com\\\/gif.latex?\\\\frac{\\\\frac{1}{\\\\sqrt{2}}}{\\\\frac{2}{\\\\sqrt{3}}+2}\",\"datePublished\":\"2023-09-08T04:36:25+00:00\",\"description\":\"NCERT Solutions Class 10 Maths\u00a0Chapter - 8 (Introduction to Trigonometry)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 8 Introduction to Trigonometry\u00a0 Exercise 8.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 8.2\",\"breadcrumb\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/#primaryimage\",\"url\":\"https:\\\/\\\/latex.codecogs.com\\\/gif.latex?\\\\frac{\\\\frac{1}{\\\\sqrt{2}}}{\\\\frac{2}{\\\\sqrt{3}}+2}\",\"contentUrl\":\"https:\\\/\\\/latex.codecogs.com\\\/gif.latex?\\\\frac{\\\\frac{1}{\\\\sqrt{2}}}{\\\\frac{2}{\\\\sqrt{3}}+2}\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\\\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#website\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/\",\"name\":\"TheExamPillar NCERT\",\"description\":\"\",\"publisher\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"alternateName\":\"NCERT Solution\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":[\"Person\",\"Organization\"],\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\",\"name\":\"Admin\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"contentUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"width\":512,\"height\":512,\"caption\":\"Admin\"},\"logo\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\"},\"sameAs\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\"],\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/author\\\/ncert_eng_vikram\\\/\"}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 | TheExamPillar NCERT","description":"NCERT Solutions Class 10 Maths\u00a0Chapter - 8 (Introduction to Trigonometry)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 8 Introduction to Trigonometry\u00a0 Exercise 8.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 8.2","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/","og_locale":"en_US","og_type":"article","og_title":"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2","og_description":"NCERT Solutions Class 10 Maths\u00a0Chapter - 8 (Introduction to Trigonometry)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 8 Introduction to Trigonometry\u00a0 Exercise 8.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 8.2","og_url":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/","og_site_name":"TheExamPillar NCERT","article_published_time":"2023-09-08T04:36:25+00:00","og_image":[{"url":"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{1}{\\sqrt{2}}}{\\frac{2}{\\sqrt{3}}+2}","type":"","width":"","height":""}],"author":"Admin","twitter_card":"summary_large_image","twitter_creator":"@exampillar","twitter_site":"@exampillar","twitter_misc":{"Written by":"Admin","Est. reading time":"8 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/#article","isPartOf":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/"},"author":{"name":"Admin","@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"headline":"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2","datePublished":"2023-09-08T04:36:25+00:00","mainEntityOfPage":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/"},"wordCount":594,"commentCount":0,"publisher":{"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"image":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/#primaryimage"},"thumbnailUrl":"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{1}{\\sqrt{2}}}{\\frac{2}{\\sqrt{3}}+2}","keywords":["Class 10 NCERT Mathematics Solutions","NCERT Class 10 Mathematics\u00a0Chapter 8 Introduction to Trigonometry Solutions","NCERT Class 10 Mathematics Exercise 8.2 Solutions","NCERT Class 10 Mathematics Solutions Class 10 NCERT Solutions","NCERT Solutions Class 10 Mathematics","NCERT Solutions Class 10 Maths"],"articleSection":["Class 10 Maths"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/","url":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/","name":"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 | TheExamPillar NCERT","isPartOf":{"@id":"https:\/\/theexampillar.com\/ncert\/#website"},"primaryImageOfPage":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/#primaryimage"},"image":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/#primaryimage"},"thumbnailUrl":"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{1}{\\sqrt{2}}}{\\frac{2}{\\sqrt{3}}+2}","datePublished":"2023-09-08T04:36:25+00:00","description":"NCERT Solutions Class 10 Maths\u00a0Chapter - 8 (Introduction to Trigonometry)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 8 Introduction to Trigonometry\u00a0 Exercise 8.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 8.2","breadcrumb":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/#primaryimage","url":"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{1}{\\sqrt{2}}}{\\frac{2}{\\sqrt{3}}+2}","contentUrl":"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{1}{\\sqrt{2}}}{\\frac{2}{\\sqrt{3}}+2}"},{"@type":"BreadcrumbList","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/theexampillar.com\/ncert\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2"}]},{"@type":"WebSite","@id":"https:\/\/theexampillar.com\/ncert\/#website","url":"https:\/\/theexampillar.com\/ncert\/","name":"TheExamPillar NCERT","description":"","publisher":{"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"alternateName":"NCERT Solution","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/theexampillar.com\/ncert\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":["Person","Organization"],"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1","name":"Admin","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","url":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","contentUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","width":512,"height":512,"caption":"Admin"},"logo":{"@id":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png"},"sameAs":["https:\/\/theexampillar.com\/ncert"],"url":"https:\/\/theexampillar.com\/ncert\/author\/ncert_eng_vikram\/"}]}},"_links":{"self":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/6070","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/comments?post=6070"}],"version-history":[{"count":3,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/6070\/revisions"}],"predecessor-version":[{"id":6531,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/6070\/revisions\/6531"}],"wp:attachment":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/media?parent=6070"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/categories?post=6070"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/tags?post=6070"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}