{"id":6072,"date":"2023-09-09T07:40:14","date_gmt":"2023-09-09T07:40:14","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6072"},"modified":"2023-09-09T07:40:14","modified_gmt":"2023-09-09T07:40:14","slug":"ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-4\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4"},"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 8.4 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-8-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 8.2<\/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<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 8.4<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong>cosec<sup>2<\/sup>A<sup>\u00a0<\/sup>&#8211; cot<sup>2<\/sup>A = 1<br \/>\n\u21d2 cosec<sup>2<\/sup>A = 1\u00a0+ cot<sup>2<\/sup>A<br \/>\n\u21d2 1\/sin<sup>2<\/sup>A = 1\u00a0+ cot<sup>2<\/sup>A<br \/>\n\u21d2 sin<sup>2<\/sup>A = 1\/(1+cot<sup>2<\/sup>A)<br \/>\n\u21d2 sin A = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\pm&amp;space;1}{\\sqrt{1+cot^2A}}\" alt=\"\\frac{\\pm 1}{\\sqrt{1+cot^2A}}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">Now,<br \/>\nsin<sup>2<\/sup>A = 1\/(1+cot<sup>2<\/sup>A)<br \/>\n\u21d2 1 &#8211; cos<sup>2<\/sup>A = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{(1+cot^2A)}\" alt=\"\\frac{1}{(1+cot^2A)}\" align=\"absmiddle\" \/><br \/>\n\u21d2 cos<sup>2<\/sup>A = 1 &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{(1+cot^2A)}\" alt=\"\\frac{1}{(1+cot^2A)}\" align=\"absmiddle\" \/><br \/>\n\u21d2 cos<sup>2<\/sup>A = (1 + cot<sup>2<\/sup>A -1)\/(1 + cot<sup>2<\/sup>A)<br \/>\n\u21d2 1\/sec<sup>2<\/sup>A = cot<sup>2<\/sup>A\/(1+cot<sup>2<\/sup>A)<br \/>\n\u21d2 sec A = (1 + cot<sup>2<\/sup>A)\/cot<sup>2<\/sup>A<br \/>\n\u21d2 sec A =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\pm&amp;space;\\sqrt{1+cot^2A}}{cotA}\" alt=\"\\frac{\\pm \\sqrt{1+cot^2A}}{cotA}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">also,<br \/>\ntan A = sin A\/cos A<br \/>\ncot A = cos A\/sin A<br \/>\n\u21d2 tan A = 1\/cot A<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. Write all the other trigonometric ratios of \u2220A in terms of sec A.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> \u00a0We know that,<br \/>\nsec A = 1\/cos A<br \/>\n\u21d2 cos A = 1\/sec A<br \/>\nalso,<br \/>\ncos<sup>2<\/sup>A\u00a0+ sin<sup>2<\/sup>A = 1<br \/>\n\u21d2 sin<sup>2<\/sup>A = 1 &#8211; cos<sup>2<\/sup>A<br \/>\n\u21d2 sin<sup>2<\/sup>A = 1 &#8211; (1\/sec<sup>2<\/sup>A)<br \/>\n\u21d2 sin<sup>2<\/sup>A = (sec<sup>2<\/sup>A &#8211; 1)\/sec<sup>2<\/sup>A<br \/>\n\u21d2 sin A = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\pm&amp;space;\\sqrt{sec^2A&amp;space;-1}}{secA}\" alt=\"\\frac{\\pm \\sqrt{sec^2A -1}}{secA}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">also,<br \/>\nsin A = 1\/cosec A<br \/>\n\u21d2 cosec A = 1\/sin A<br \/>\n\u21d2 cosec A = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\pm&amp;space;secA}{\\sqrt{sec^2A-1}}\" alt=\"\\frac{\\pm secA}{\\sqrt{sec^2A-1}}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">Now,<br \/>\nsec<sup>2<\/sup>A &#8211; tan<sup>2<\/sup>A = 1<br \/>\n\u21d2 tan<sup>2<\/sup>A = sec<sup>2<\/sup>A\u00a0+ 1<br \/>\n\u21d2 tan A = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{sec^2A&amp;space;+1}\" alt=\"\\sqrt{sec^2A +1}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">also,<br \/>\ntan A = 1\/cot A<br \/>\n\u21d2 cot A = 1\/tan A<br \/>\n\u21d2 cot A = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\pm1}{\\sqrt{sec^2A&amp;space;+1}}\" alt=\"\\frac{\\pm1}{\\sqrt{sec^2A +1}}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3.\u00a0Evaluate:<br \/>\n<\/strong><strong>(i) (sin<sup>2<\/sup>63\u00b0 + sin<sup>2<\/sup>27\u00b0)\/(cos<sup>2<\/sup>17\u00b0 + cos<sup>2<\/sup>73\u00b0)<br \/>\n(ii)\u00a0\u00a0sin 25\u00b0 cos 65\u00b0 + cos 25\u00b0 sin 65\u00b0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) (sin<sup>2<\/sup>63\u00b0 + sin<sup>2<\/sup>27\u00b0)\/(cos<sup>2<\/sup>17\u00b0 + cos<sup>2<\/sup>73\u00b0)<br \/>\n<\/strong>= (sin<sup>2<\/sup>(90\u00b0 &#8211; 27\u00b0) + sin<sup>2<\/sup>27\u00b0) \/ (cos<sup>2<\/sup>(90\u00b0 &#8211; 73\u00b0) + cos<sup>2<\/sup>73\u00b0))<br \/>\n= (cos<sup>2<\/sup>27\u00b0<sup>\u00a0<\/sup>+ sin<sup>2<\/sup>27\u00b0)\/(sin<sup>2<\/sup>27\u00b0\u00a0+ cos<sup>2<\/sup>73\u00b0)<br \/>\n= 1\/1 (\u2235 sin<sup>2<\/sup>A\u00a0+ cos<sup>2<\/sup>A = 1)<br \/>\n= 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) sin 25\u00b0 cos 65\u00b0 + cos 25\u00b0 sin 65\u00b0<br \/>\n<\/strong>= sin(90\u00b0 &#8211; 25\u00b0) cos 65\u00b0 + cos(90\u00b0 &#8211; 65\u00b0) sin 65\u00b0<br \/>\n= cos 65\u00b0 cos 65\u00b0 + sin 65\u00b0 sin 65\u00b0<br \/>\n= cos<sup>2<\/sup>65\u00b0<sup>\u00a0<\/sup>+ sin<sup>2<\/sup>65\u00b0<br \/>\n= 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. Choose the correct option. Justify your choice.<br \/>\n(i) 9 sec<sup>2<\/sup>A \u2013 9 tan<sup>2<\/sup>A =<br \/>\n<\/strong>(A) 1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0(B) 9 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C) 8 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(D) 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) (1 + tan \u03b8 + sec \u03b8) (1 + cot \u03b8 \u2013 cosec \u03b8)<br \/>\n<\/strong>(A) 0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (B) 1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C) 2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(D) \u2013 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) (sec A + tan A) (1 \u2013 sin A) =<br \/>\n<\/strong>(A) sec A \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0(B) sin A \u00a0 \u00a0 \u00a0 \u00a0(C) cosec A \u00a0 \u00a0 \u00a0(D) cos A<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) 1+tan<sup>2<\/sup>A\/1+cot<sup>2<\/sup>A =<br \/>\n<\/strong>(A) sec<sup>2 <\/sup>A\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(B) -1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C)\u00a0cot<sup>2<\/sup>A\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0(D) tan<sup>2<\/sup>A<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) 9 sec<sup>2<\/sup>A \u2013 9 tan<sup>2<\/sup>A<br \/>\n<\/strong>= 9 (sec<sup>2<\/sup>A\u00a0&#8211; tan<sup>2<\/sup>A)<br \/>\n= 9 \u00d7 1 (\u2235 sec2 A &#8211; tan2 A = 1)<br \/>\n= 9<br \/>\n<strong>(B) is correct.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) (1 + tan \u03b8 + sec \u03b8) (1 + cot \u03b8 &#8211; cosec \u03b8)<br \/>\n<\/strong>= (1\u00a0+ sin \u03b8\/cos \u03b8\u00a0+ 1\/cos \u03b8) (1\u00a0+ cos \u03b8\/sin \u03b8 &#8211; 1\/sin \u03b8)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (cos \u03b8 + sin \u03b8 + 1)\/cos \u03b8 \u00d7 (sin \u03b8 + cos \u03b8 &#8211; 1)\/sin \u03b8<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (cos \u03b8 + sin \u03b8)<sup>2 <\/sup>&#8211; 1<sup>2<\/sup>\/(cos \u03b8 sin \u03b8)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (cos<sup>2<\/sup>\u03b8 + sin<sup>2<\/sup>\u03b8 + 2cos \u03b8 sin \u03b8 &#8211; 1)\/(cos \u03b8 sin \u03b8)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (1 + 2cos \u03b8 sin \u03b8 &#8211; 1)\/(cos \u03b8 sin \u03b8)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (2cos \u03b8 sin \u03b8)\/(cos \u03b8 sin \u03b8) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(C) is correct<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) (secA + tanA) (1 &#8211; sinA)<br \/>\n<\/strong>= (1\/cos A\u00a0+ sin A\/cos A) (1 &#8211; sinA)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (1 + sin A\/cos A) (1 &#8211; sinA)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (1 &#8211; sin<sup>2<\/sup>A)\/cos A<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= cos<sup>2<\/sup>A\/cos A <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= cos A <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(D) is correct.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) (1 + tan<sup>2<\/sup>A)\/(1 + cot<sup>2<\/sup>A)<\/strong>\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (1 + 1\/cot<sup>2<\/sup>A)\/(1 + cot<sup>2<\/sup>A)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (cot<sup>2<\/sup>A + 1\/cot<sup>2<\/sup>A) \/(1 + cot<sup>2<\/sup>A)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">=\u00a01\/cot<sup>2<\/sup>A <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= tan<sup>2<\/sup>A<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(D) is correct.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5.\u00a0Prove the following identities, where the angles involved are acute angles for which the<br \/>\nexpressions are defined.<br \/>\n<\/strong>(i) (cosec \u03b8 \u2013 cot \u03b8)<sup>2\u00a0<\/sup>= (1-cos \u03b8)\/(1+cos \u03b8)<br \/>\n(ii) <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{coaA}{1+sinA}&amp;space;+&amp;space;\\frac{1+sinA}{cosA}\" alt=\"\\frac{coaA}{1+sinA} + \\frac{1+sinA}{cosA}\" align=\"absmiddle\" \/><\/strong>= 2 sec A<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">(iii) <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{tan\\theta}{1-cot\\theta}&amp;space;+\\frac{cot\\theta}{1-tan\\theta}\" alt=\"\\frac{tan\\theta}{1-cot\\theta} +\\frac{cot\\theta}{1-tan\\theta}\" align=\"absmiddle\" \/> <\/strong>= 1 + sec \u03b8 cosec \u03b8 [<strong>Hint :<\/strong> Write the expression in terms of sin \u03b8 and cos \u03b8]<br \/>\n(iv) (1\u00a0+ sec A)\/sec A = sin<sup>2<\/sup>A\/(1-cos A) \u00a0[<strong>Hint :<\/strong> Simplify LHS and RHS separately]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">(v) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{cosA&amp;space;-&amp;space;sin&amp;space;A&amp;space;+&amp;space;1}{cos&amp;space;A+sinA-1}\" alt=\"\\frac{cosA - sin A + 1}{cos A+sinA-1}\" align=\"absmiddle\" \/> = cosec A + cot A, using the identity cosec<sup>2<\/sup>A = 1 + cot<sup>2<\/sup>A.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">(vi) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{\\frac{1+sinA}{1-sinA}}&amp;space;=&amp;space;secA&amp;space;+&amp;space;tanA\" alt=\"\\sqrt{\\frac{1+sinA}{1-sinA}} = secA + tanA\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">(vii) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(sin\\theta-2sin^3\\theta)}{2cos^3\\theta-cos\\theta}\" alt=\"\\frac{(sin\\theta-2sin^3\\theta)}{2cos^3\\theta-cos\\theta}\" align=\"absmiddle\" \/> = tan \u03b8<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">(viii) (sin A\u00a0+ cosec A)<sup>2\u00a0<\/sup>+ (cos A + sec A)<sup>2<\/sup> = 7 + tan<sup>2<\/sup>A + cot<sup>2<\/sup>A <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">(ix) (cosec A \u2013 sin A)(sec A \u2013 cos A) = 1\/(tan A + cotA) [<strong>Hint :<\/strong> Simplify LHS and RHS separately]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">(x) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{1+tan^2A}{1+cot^2A}&amp;space;\\right&amp;space;)&amp;space;=&amp;space;\\left&amp;space;(&amp;space;\\frac{1-tanA}{1-cotA}&amp;space;\\right&amp;space;)^2&amp;space;=&amp;space;tan^2A\" alt=\"\\left ( \\frac{1+tan^2A}{1+cot^2A} \\right ) = \\left ( \\frac{1-tanA}{1-cotA} \\right )^2 = tan^2A\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) (cosec \u03b8 \u2013 cot \u03b8)<sup>2\u00a0<\/sup>= (1 &#8211; cos \u03b8)\/(1 + cos \u03b8)<br \/>\n<\/strong>L.H.S. = (cosec \u03b8 &#8211; cot \u03b8)<sup>2<br \/>\n<\/sup>= (cosec<sup>2<\/sup>\u03b8 +\u00a0cot<sup>2<\/sup>\u03b8 &#8211; 2cosec \u03b8 cot \u03b8)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (1\/sin<sup>2<\/sup>\u03b8\u00a0+ cos<sup>2<\/sup>\u03b8\/sin<sup>2<\/sup>\u03b8 &#8211; 2cos \u03b8\/sin<sup>2<\/sup>\u03b8)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (1 + cos<sup>2<\/sup>\u03b8 &#8211; 2cos \u03b8)\/(1 &#8211; cos<sup>2<\/sup>\u03b8)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (1 &#8211; cos \u03b8)<sup>2<\/sup>\/(1 &#8211; cos\u03b8)(1 + cos \u03b8)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (1 &#8211; cos \u03b8)\/(1 + cos \u03b8) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= R.H.S.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{coaA}{1+sinA}&amp;space;+&amp;space;\\frac{1+sinA}{cosA}\" alt=\"\\frac{coaA}{1+sinA} + \\frac{1+sinA}{cosA}\" align=\"absmiddle\" \/>= 2 sec A<br \/>\n<\/strong>L.H.S. = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{coaA}{1+sinA}&amp;space;+&amp;space;\\frac{1+sinA}{cosA}\" alt=\"\\frac{coaA}{1+sinA} + \\frac{1+sinA}{cosA}\" align=\"absmiddle\" \/><br \/>\n<\/strong>= [cos<sup>2<\/sup>A + (1 + sin A)<sup>2<\/sup>]\/(1 + sin A) cos A<br \/>\n= (cos<sup>2<\/sup>A + sin<sup>2<\/sup>A + 1 + 2sin A)\/(1 + sin A) cos A<br \/>\nSince cos<sup>2<\/sup>A + sin<sup>2<\/sup>A = 1<br \/>\n= (1 + 1 + 2sin A)\/(1 + sin A) cos A<br \/>\n= (2+ 2sin A)\/(1 + sin A) cos A<br \/>\n= 2(1 + sin A)\/(1 + sin A)cos A<br \/>\n= 2\/cos A<br \/>\n= 2 sec A<br \/>\n= R.H.S.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{tan\\theta}{1-cot\\theta}&amp;space;+\\frac{cot\\theta}{1-tan\\theta}\" alt=\"\\frac{tan\\theta}{1-cot\\theta} +\\frac{cot\\theta}{1-tan\\theta}\" align=\"absmiddle\" \/> = 1 + sec \u03b8 cosec \u03b8 <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">L.H.S. = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{tan\\theta}{1-cot\\theta}&amp;space;+\\frac{cot\\theta}{1-tan\\theta}\" alt=\"\\frac{tan\\theta}{1-cot\\theta} +\\frac{cot\\theta}{1-tan\\theta}\" align=\"absmiddle\" \/> \u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">We know that tan \u03b8 = sin \u03b8\/cos \u03b8<br \/>\ncot \u03b8 = cos \u03b8\/sin \u03b8<br \/>\n= [(sin \u03b8\/cos \u03b8)\/1 &#8211; (cos \u03b8\/sin \u03b8)] + [(cos \u03b8\/sin \u03b8)\/1 &#8211; (sin \u03b8\/cos \u03b8)]<br \/>\n= [(sin \u03b8\/cos \u03b8)\/(sin \u03b8 &#8211; cos \u03b8)\/sin \u03b8] + [(cos \u03b8\/sin \u03b8)\/(cos \u03b8 &#8211; sin \u03b8)\/cos \u03b8]<br \/>\n= sin<sup>2<\/sup>\u03b8\/[cos \u03b8(sin \u03b8 &#8211; cos \u03b8)] + cos<sup>2<\/sup>\u03b8\/[sin \u03b8(cos \u03b8 &#8211; sin \u03b8)]<br \/>\n= sin<sup>2<\/sup>\u03b8\/[cos \u03b8(sin \u03b8 &#8211; cos \u03b8)] \u2013 cos<sup>2<\/sup>\u03b8\/[sin \u03b8(sin \u03b8 &#8211; cos \u03b8)]<br \/>\n= 1\/(sin \u03b8 &#8211; cos \u03b8) [(sin<sup>2<\/sup>\u03b8\/cos \u03b8) \u2013 (cos<sup>2<\/sup>\u03b8\/sin \u03b8)]<br \/>\n= 1\/(sin \u03b8 &#8211; cos \u03b8) \u00d7 [(sin<sup>3<\/sup>\u03b8 \u2013 cos<sup>3<\/sup>\u03b8)\/sin\u00a0\u03b8 cos \u03b8]<br \/>\n= [(sin \u03b8 &#8211; cos \u03b8)(sin<sup>2<\/sup>\u03b8 + cos<sup>2<\/sup>\u03b8 + sin \u03b8 cos \u03b8)]\/[(sin \u03b8 &#8211; cos \u03b8)sin \u03b8 cos \u03b8]<br \/>\n= (1\u00a0+ sin\u00a0\u03b8 cos \u03b8)\/sin\u00a0\u03b8 cos \u03b8<br \/>\n= 1\/sin\u00a0\u03b8 cos \u03b8\u00a0+ 1<br \/>\n= 1\u00a0+ sec \u03b8 cosec \u03b8<br \/>\n= R.H.S.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1+secA}{secA}&amp;space;=&amp;space;\\frac{sin^2A}{1-cosA}\" alt=\"\\frac{1+secA}{secA} = \\frac{sin^2A}{1-cosA}\" width=\"167\" height=\"41\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">First find the simplified form of L.H.S<br \/>\nL.H.S. = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1+secA}{secA}\" alt=\"\\frac{1+secA}{secA}\" align=\"absmiddle\" \/><br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1+\\frac{1}{cosA}}{\\frac{1}{cosA}}\" alt=\"\\frac{1+\\frac{1}{cosA}}{\\frac{1}{cosA}}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{cosA+1}{cosA}}{\\frac{1}{cosA}}\" alt=\"\\frac{\\frac{cosA+1}{cosA}}{\\frac{1}{cosA}}\" align=\"absmiddle\" \/><br \/>\n= cos A + 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">R.H.S. = sin<sup>2<\/sup>A\/(1-cos A)<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(1-cos^2A)}{1-cosA}\" alt=\"\\frac{(1-cos^2A)}{1-cosA}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">= <span style=\"text-align: justify;\">(1 &#8211; cos A)(1 + cos A)\/(1 &#8211; cos A)<br \/>\n<\/span>= cos A + 1<br \/>\nL.H.S. = R.H.S.<br \/>\nHence proved<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">(v) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{cosA&amp;space;-&amp;space;sin&amp;space;A&amp;space;+&amp;space;1}{cos&amp;space;A+sinA-1}\" alt=\"\\frac{cosA - sin A + 1}{cos A+sinA-1}\" align=\"absmiddle\" \/> = cosec A + cot A<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">L.H.S. = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{cosA&amp;space;-&amp;space;sin&amp;space;A&amp;space;+&amp;space;1}{cos&amp;space;A+sinA-1}\" alt=\"\\frac{cosA - sin A + 1}{cos A+sinA-1}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{cosA&amp;space;-&amp;space;sin&amp;space;A&amp;space;+&amp;space;1}{sinA}}{\\frac{cos&amp;space;A+sinA-1}{sinA}}\" alt=\"\\frac{\\frac{cosA - sin A + 1}{sinA}}{\\frac{cos A+sinA-1}{sinA}}\" align=\"absmiddle\" \/><br \/>\nWe know that cos A\/sin A = cot A and 1\/sin A = cosec A<br \/>\n= (cot A \u2013 1 + cosec A)\/(cot A + 1 \u2013 cosec A)<br \/>\n= (cot A \u2013 cosec<sup>2<\/sup>A\u00a0+ cot<sup>2<\/sup>A + cosec A)\/(cot A + 1 \u2013 cosec A)\u00a0 \u00a0 \u00a0(using cosec<sup>2<\/sup>A \u2013 cot<sup>2<\/sup>A = 1)<br \/>\n= ((cot A + cosec A) \u2013 (cosec<sup>2<\/sup>A\u00a0\u2013 cot<sup>2<\/sup>A))\/(cot A + 1 \u2013 cosec A)<br \/>\n= ((cot A + cosec A) \u2013 (cosec A + cot A)(cosec A \u2013 cot A))\/(1 \u2013 cosec A + cot A)<br \/>\n= \u00a0(cot A\u00a0+\u00a0cosec A)(1 \u2013 cosec A\u00a0+\u00a0cot A)\/(1 \u2013 cosec A\u00a0+\u00a0cot A)<br \/>\n= \u00a0cot A\u00a0+\u00a0cosec A<br \/>\n= R.H.S.<br \/>\nHence Proved<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">(vi) <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{\\frac{1+sinA}{1-sinA}}&amp;space;=&amp;space;secA&amp;space;+&amp;space;tanA\" alt=\"\\sqrt{\\frac{1+sinA}{1-sinA}} = secA + tanA\" width=\"213\" height=\"44\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">L.H.S. = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{\\frac{1+sinA}{1-sinA}}\" alt=\"\\sqrt{\\frac{1+sinA}{1-sinA}}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{\\frac{\\frac{1+sinA}{cosA}}{\\frac{1-sinA}{cosA}}}\" alt=\"\\sqrt{\\frac{\\frac{1+sinA}{cosA}}{\\frac{1-sinA}{cosA}}}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">We know that 1\/cos A = sec A and sin A\/cos A = tan A<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{\\frac{secA+tanA}{secA-tanA}}\" alt=\"\\sqrt{\\frac{secA+tanA}{secA-tanA}}\" align=\"absmiddle\" \/><br \/>\nNow using rationalization, we get<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">= <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{\\frac{secA+tanA}{secA-tanA}}\" alt=\"\\sqrt{\\frac{secA+tanA}{secA-tanA}}\" width=\"120\" height=\"44\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{\\frac{secA&amp;space;+&amp;space;tanA}{secA&amp;space;+&amp;space;tanA}}\" alt=\"\\sqrt{\\frac{secA + tanA}{secA + tanA}}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{\\frac{(secA+tanA)^2}{sec^2A-tan^2A}}\" alt=\"\\sqrt{\\frac{(secA+tanA)^2}{sec^2A-tan^2A}}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">= (sec A + tan A)\/1<br \/>\n= sec A + tan A<br \/>\n= R.H.S<br \/>\nHence proved<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(vii)<\/strong> <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(sin\\theta-2sin^3\\theta)}{2cos^3\\theta-cos\\theta}\" alt=\"\\frac{(sin\\theta-2sin^3\\theta)}{2cos^3\\theta-cos\\theta}\" align=\"absmiddle\" \/> = tan \u03b8<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">L.H.S. = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(sin\\theta-2sin^3\\theta)}{2cos^3\\theta-cos\\theta}\" alt=\"\\frac{(sin\\theta-2sin^3\\theta)}{2cos^3\\theta-cos\\theta}\" align=\"absmiddle\" \/><br \/>\n= (sin \u03b8 (1 \u2013 2sin<sup>2<\/sup>\u03b8))\/(cos \u03b8(2cos<sup>2<\/sup>\u03b8 &#8211; 1))<br \/>\nWe know that sin<sup>2<\/sup>\u03b8 = 1 &#8211; cos<sup>2<\/sup>\u03b8<br \/>\n= (sin \u03b8(1 \u2013 2(1 &#8211; cos<sup>2<\/sup>\u03b8))\/(cos \u03b8(2cos<sup>2<\/sup>\u03b8 &#8211; 1))<br \/>\n= (sin \u03b8(2cos<sup>2<\/sup>\u03b8 &#8211; 1))\/(cos \u03b8(2cos<sup>2<\/sup>\u03b8 -1))<br \/>\n= sin \u03b8\/ cos \u03b8<br \/>\n= tan \u03b8<br \/>\n= R.H.S.<br \/>\nHence proved<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(viii) (sin A\u00a0+ cosec A)<sup>2\u00a0<\/sup>+ (cos A + sec A)<sup>2<\/sup>\u00a0=\u00a07+tan<sup>2<\/sup>A+cot<sup>2<\/sup>A<br \/>\n<\/strong>L.H.S. = (sin A\u00a0+ cosec A)<sup>2\u00a0<\/sup>+ (cos A + sec A)<sup>2<br \/>\n<\/sup>(a+b)<sup>2<\/sup> = a<sup>2<\/sup> + b<sup>2<\/sup> +2ab<br \/>\n= (sin<sup>2<\/sup>A\u00a0+ cosec<sup>2<\/sup>A\u00a0+ 2 sin A cosec A)\u00a0+ (cos<sup>2<\/sup>A\u00a0+\u00a0sec<sup>2<\/sup>A\u00a0+ 2 cos A sec A)<br \/>\n= (sin<sup>2<\/sup>A + cos<sup>2<\/sup>A) + 2 sin A(1\/sin A)\u00a0+ 2 cos A(1\/cos A)\u00a0+ 1 + tan<sup>2<\/sup>A + 1\u00a0+ cot<sup>2<\/sup>A<br \/>\n= 1\u00a0+ 2\u00a0+ 2\u00a0+ 2 + tan<sup>2<\/sup>A\u00a0+\u00a0cot<sup>2<\/sup>A<br \/>\n= 7 + tan<sup>2<\/sup>A + cot<sup>2<\/sup>A<br \/>\n= R.H.S.<br \/>\nHence proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ix) (cosec A \u2013 sin A)(sec A \u2013 cos A) = 1\/(tan A+cotA)<br \/>\n<\/strong>L.H.S. = (cosec A \u2013 sin A)(sec A \u2013 cos A)<br \/>\n= (1\/sin A \u2013 sin A)(1\/cos A \u2013 cos A)<br \/>\n= ((1 &#8211; sin<sup>2<\/sup>A)\/sin A][(1 &#8211; cos<sup>2<\/sup>A)\/cos A)<br \/>\n= (cos<sup>2<\/sup>A\/sin A) \u00d7 (sin<sup>2<\/sup>A\/cos A)<br \/>\n= cos A sin A<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">R.H.S. = 1\/(tan A + cotA)<br \/>\n= 1\/(sin A\/cos A\u00a0 + cos A\/sin A)<br \/>\n= 1\/[(sin<sup>2<\/sup>A + cos<sup>2<\/sup>A)\/sin A cos A]<br \/>\n= cos A sin A <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">L.H.S. = R.H.S.<br \/>\nHence proved<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(x)<\/strong> <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{1+tan^2A}{1+cot^2A}&amp;space;\\right&amp;space;)&amp;space;=&amp;space;\\left&amp;space;(&amp;space;\\frac{1-tanA}{1-cotA}&amp;space;\\right&amp;space;)^2&amp;space;=&amp;space;tan^2A\" alt=\"\\left ( \\frac{1+tan^2A}{1+cot^2A} \\right ) = \\left ( \\frac{1-tanA}{1-cotA} \\right )^2 = tan^2A\" width=\"310\" height=\"48\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">L.H.S. = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{1+tan^2A}{1+cot^2A}&amp;space;\\right&amp;space;)\" alt=\"\\left ( \\frac{1+tan^2A}{1+cot^2A} \\right )\" align=\"absmiddle\" \/><br \/>\n= (1 + tan<sup>2<\/sup>A)\/(1 + 1\/tan<sup>2<\/sup>A))<br \/>\n= (1 + tan<sup>2<\/sup>A)\/((1 + tan<sup>2<\/sup>A)\/tan<sup>2<\/sup>A)<br \/>\n= tan<sup>2<\/sup>A<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">Similarly,<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{1-tanA}{1-cotA}&amp;space;\\right&amp;space;)^2\" alt=\"\\left ( \\frac{1-tanA}{1-cotA} \\right )^2\" align=\"absmiddle\" \/> =<sup>\u00a0<\/sup>tan<sup>2<\/sup>A<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence proved<\/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 8.4 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 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