{"id":13073,"date":"2022-06-05T22:30:25","date_gmt":"2022-06-05T17:00:25","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=13073"},"modified":"2022-05-23T16:10:38","modified_gmt":"2022-05-23T10:40:38","slug":"introduction-to-trigonometry-class-10-extra-questions","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/introduction-to-trigonometry-class-10-extra-questions\/","title":{"rendered":"Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers"},"content":{"rendered":"

Here you will find Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Answers Solutions, Extra Questions for Class 10 Maths<\/a> are solved by experts and will guide students in the right direction.<\/p>\n

Extra Questions for Class 10 Maths Introduction to Trigonometry with Answers Solutions<\/h2>\n

Extra Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry with Solutions Answers<\/strong><\/p>\n

Introduction to Trigonometry Class 10 Extra Questions Objective Type<\/h3>\n

Question 1.
\nThe value of \\(\\frac{2 \\tan 30^{\\circ}}{1-\\tan ^{2} 30^{\\circ}}\\) will be:
\n(i) sin 60\u00b0
\n(ii) cos 60\u00b0
\n(iii) tan 60\u00b0
\n(iv) sin 30\u00b0
\nAnswer:
\n(ii) cos 60\u00b0
\nSolution.
\n\"Introduction
\n= 2 sin 30o. cos 30\u00b0
\n= sin 60\u00b0
\nHence, Choice (ii) is correct<\/p>\n

Question 2.
\nIf sin \u03b8 = 1, the value of sin 20 will be :
\n(i) -1
\n(ii) 0
\n(iii) 1
\n(iv) 2
\nAnswer:
\n(ii) 0<\/p>\n

Question 3.
\nThe value of cot (- 1470\u00b0) will be :
\n(i) –\\(\\frac{1}{\\sqrt{3}}\\)
\n(ii) – \u221a3
\n(iii) \\(\\frac{1}{\\sqrt{3}}\\)
\n(iv) \u221a3
\nAnswer:
\n(ii) – \u221a3<\/p>\n

Question 4.
\nIn triangle ABC, the value of sin (B + C) will be :
\n(i) sin B + sin C
\n(ii) sin A
\n(iii) 0
\n(iv) cos A
\nAnswer:
\n(ii) sin A<\/p>\n

\"Introduction<\/p>\n

Question 5.
\nThe value of cos 15\u00b0 is :
\n(i) \"Introduction
\n(ii) \"Introduction
\n(iii) \"Introduction
\n(iv) \"Introduction
\nAnswer:
\n(i) \"Introduction<\/p>\n

Question 6.
\nIf sin \u03b8 = cosec \u03b8 and \u2264 \u03b8 \u2264 \u03c0 then value of 0 will be :
\n(i) \u03c0
\n(ii) \\(\\frac{\\pi}{2}\\)
\n(iii) \\(\\frac{\\pi}{4}\\)
\n(iv) 0
\nAnswer:
\n(ii) \\(\\frac{\\pi}{2}\\)<\/p>\n

Question 7.
\nThe value of cosec 810 will be :
\n(i) -1
\n(ii) 0
\n(iii) 1
\n(iv) \u221e
\nAnswer:
\n(iii) 1<\/p>\n

Question 8.
\nFind the value of \\(\\frac{\\sin ^{2} 15^{\\circ}-\\cos ^{2} 15^{\\circ}}{\\sin ^{2} 15^{\\circ}+\\cos ^{2} 15^{\\circ}}\\).
\n(i) 1
\n(ii) -1
\n(iii) \\(\\frac{2}{\\sqrt{3}}\\)
\n(iv) \\(\\frac{-\\sqrt{3}}{2}\\)
\nAnswer:
\n(iv) \\(\\frac{-\\sqrt{3}}{2}\\)<\/p>\n

Question 9.
\nThe value of \"Introduction will be :
\n(i) \\(\\frac{1}{\\sqrt{2}}\\)
\n(ii) \u221a2 – 1
\n(iii) \u221a2
\n(iv) \\(\\frac{\\sqrt{2}+1}{\\sqrt{2}-1}\\)
\nAnswer:
\n(iii) \u221a2<\/p>\n

\"Introduction<\/p>\n

Question 10.
\nThe value of sin 570\u00b0 will be :
\n(i) \\(\\frac{-1}{2}\\)
\n(ii) \\(\\frac{-\\sqrt{3}}{2}\\)
\n(iii) \\(\\frac{1}{2}\\)
\n(iv) \\(\\frac{\\sqrt{3}}{2}\\)
\nAnswer:
\n(i) \\(\\frac{-1}{2}\\)<\/p>\n

Question 11.
\nIf tan A = \\(\\frac{1}{\\sqrt{3}}\\) and tan B = \u221a3 then the value of tan (A + B) will be :
\n(i) 0
\n(ii) \\(\\frac{1}{\\sqrt{3}}\\)
\n(iii) 1
\n(iv) \u221e
\nAnswer:
\n(iv) \u221e<\/p>\n

Question 12.
\nIf sin \u03b1 = cos \u03b1 then the value of \u03b1 will be:
\n(i) 30\u00b0
\n(ii) 45\u00b0
\n(iii) 60\u00b0
\n(iv) 90\u00b0
\nAnswer:
\n(ii) 45\u00b0<\/p>\n

Question 13.
\nThe value of sin2 \u03b8 +\\(\\frac{1}{1+\\tan ^{2} \\theta}\\) is :
\n(i) cos2<\/sup> \u03b8
\n(ii) sin2<\/sup> \u03b8
\n(iii) 1
\n(iv) sec2<\/sup> \u03b8
\nAnswer:
\n(iii) 1<\/p>\n

Question 14.
\nIf sec \u03b8 = 2 then the value of \u03b8 will be :
\n(i) \\(\\frac{\\pi}{2}\\)
\n(ii) \\(\\frac{\\pi}{3}\\)
\n(iii) \\(\\frac{\\pi}{4}\\)
\n(iv) \\(\\frac{\\pi}{6}\\)
\nAnswer:
\n(ii) \\(\\frac{\\pi}{3}\\)<\/p>\n

\"Introduction<\/p>\n

Question 15.
\nThe value of cos (-1920\u00b0) will be :
\n(i) – 1\/2
\n(ii) 0
\n(iii) 1\/2
\n(iv) 1
\nAnswer:
\n(i) – 1\/2<\/p>\n

Question 16.
\nIf 2 cos 30 = 1. Then the value of a will be :
\n(i) 30\u00b0
\n(ii) 45\u00b0
\n(iii) 60\u00b0
\n(iv) 20\u00b0
\nAnswer:
\n(iv) 20\u00b0<\/p>\n

Question 17.
\nThe value of \\(\\frac{\\sin 20^{\\circ}}{\\cos 70^{\\circ}}\\) will be :
\n(i) more than 1
\n(ii) 1
\n(iii) 0
\n(iv) less than 1
\nAnswer:
\n(ii) 1<\/p>\n

Question 18.
\nThe value of sin (- 570\u00b0) is :
\n(i) –\\(\\frac{\\sqrt{3}}{2}\\)
\n(ii) \\(\\frac{\\sqrt{3}}{2}\\)
\n(iii) \\(\\frac{1}{2}\\)
\n(iv) –\\(\\frac{1}{2}\\)
\nAnswer:
\n(iii) \\(\\frac{1}{2}\\)<\/p>\n

Question 19.
\nThe value of sin 840\u00b0 will be:
\n(i) \\(\\frac{1}{2}\\)
\n(ii) –\\(\\frac{1}{2}\\)
\n(iii) \\(\\frac{\\sqrt{3}}{2}\\)
\n(iv) –\\(\\frac{\\sqrt{3}}{2}\\)
\nAnswer:
\n(iii) \\(\\frac{\\sqrt{3}}{2}\\)<\/p>\n

Question 20.
\nThe value of sin 3270\u00b0
\n(i) \\(\\frac{1}{\\sqrt{2}}\\)
\n(ii) \\(\\frac{1}{3}\\)
\n(iii) \\(\\frac{1}{2}\\)
\n(iv) \\(\\frac{\\sqrt{3}}{2}\\)
\nAnswer:
\n(iii) \\(\\frac{1}{2}\\)<\/p>\n

Question 21.
\nThe value of cos 2940\u00b0 is :
\n(i) \\(\\frac{1}{\\sqrt{3}}\\)
\n(ii) \\(\\frac{1}{2}\\)
\n(iii) \\(\\frac{1}{3}\\)
\n(iv) \\(\\frac{\\sqrt{3}}{2}\\)
\nAnswer:
\n(ii) \\(\\frac{1}{2}\\)<\/p>\n

Question 22.
\nThe value of cos2<\/sup> 61\u00b0 + cos2<\/sup> 29\u00b0 will
\n(i) 0
\n(ii) 1
\n(iii) -1
\n(iv) 2
\nAnswer:
\n(ii) 1<\/p>\n

\"Introduction<\/p>\n

Question 23.
\nIf sin \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\) and 0\u00b0 < \u03b8 < 90\u00b0, then the value of tan 2\u03b8 will be :
\n(i) -\u221a3
\n(ii) –\\(\\frac{1}{\\sqrt{3}}\\)
\n(iii) \\(\\frac{1}{\\sqrt{3}}\\)
\n(iv) \u221a3
\nAnswer:
\n(i) -\u221a3<\/p>\n

Question 24.
\nThe value of cos 2A is :
\n(i) cos2<\/sup> A – sin2<\/sup>A
\n(ii) 1 – 2 cos2<\/sup>A
\n(iii) cos2<\/sup>A + sin2<\/sup>A
\n(iv) 2 sin2<\/sup> A – 1.
\nAnswer:
\n(i) cos2<\/sup> A – sin2<\/sup>A<\/p>\n

Question 25.
\nThe palue of sin 2A is :
\n(i) \\(\\frac{1+\\tan ^{2} \\mathrm{A}}{1-\\tan ^{2} \\mathrm{A}}\\)
\n(ii) \\(\\frac{1-\\tan ^{2} \\mathrm{A}}{1+\\tan ^{2} \\mathrm{A}}\\)
\n(iii) \\(\\frac{2 \\tan \\mathrm{A}}{1-\\tan ^{2} \\mathrm{A}}\\)
\n(iv) \\(\\frac{2 \\tan \\mathrm{A}}{1+\\tan ^{2} \\mathrm{A}}\\)
\nAnswer:
\n(iv) \\(\\frac{2 \\tan \\mathrm{A}}{1+\\tan ^{2} \\mathrm{A}}\\)<\/p>\n

Question 26.
\nThe value of cos (-405\u00b0) is :
\n(i) –\\(\\frac{1}{\\sqrt{2}}\\)
\n(ii) \\(\\frac{1}{\\sqrt{2}}\\)
\n(iii) \\(\\frac{1}{2}\\)
\n(iv) –\\(\\frac{1}{2}\\)
\nAnswer:
\n(ii) \\(\\frac{1}{\\sqrt{2}}\\)<\/p>\n

Question 27.
\nIf sin \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\) and 0\u00b0 < \u03b8 < 90\u00b0, the value of cot 2\u03b8 will be :
\n(i) –\\(\\frac{1}{\\sqrt{3}}\\)
\n(ii) -\u221a3
\n(iii) \\(\\frac{1}{\\sqrt{3}}\\)
\n(iv) \u221a3
\nAnswer:
\n(i) –\\(\\frac{1}{\\sqrt{3}}\\)<\/p>\n

Question 28.
\nThe value \"Introduction will be :
\n(i) -1
\n(ii) 0
\n(iii) 1
\n(iv) 3
\nAnswer:
\n(i) -1<\/p>\n

Question 29.
\nThe value \"Introduction will be :
\n(i) -\u221e
\n(ii) -1
\n(iii) +1
\n(iv) \u221e
\nAnswer:
\n(ii) -1<\/p>\n

Question 30.
\nThe value of sec 70\u00b0 sin 20\u00b0 \u2013 cos 20\u00b0 . cosec 70\u00b0 will be :
\n(i) -1
\n(ii) 0
\n(iii) 1
\n(iv) infinity.
\nAnswer:
\n(ii) 0<\/p>\n

\"Introduction<\/p>\n

Question 31.
\nThe value of sin 12\u00b0 cos 78\u00b0 + sin 78\u00b0 cos 12\u00b0 will be:
\n(i) 2
\n(ii) 1
\n(iii) 0
\n(iv) -1.
\nAnswer:
\n(ii) 1<\/p>\n

Question 32.
\nThe maximum value of sin x for the value of x is :
\n(i) x = \\(\\frac{\\pi}{4}\\)
\n(ii) x = \\(\\frac{\\pi}{2}\\)
\n(iii) x = \u03c0
\n(iv) x = \\(\\frac{3 \\pi}{2}\\)
\nAnswer:
\n(ii) x = \\(\\frac{\\pi}{2}\\)<\/p>\n

Question 33.
\nValue of \\(\\frac{\\sin 31^{\\circ}}{\\cos 59^{\\circ}}\\) will be :
\n(i) -1
\n(ii) 0
\n(iii) 1
\n(iv) 2.
\nAnswer:
\n(iii) 1<\/p>\n

Question 34.
\nIf cos\u03b8 = \\(\\frac{1}{2}\\), the value of cosec2<\/sup>\u03b8 is :
\n(i) \\(\\frac{1}{2}\\)
\n(ii) \\(\\frac{\\sqrt{3}}{2}\\)
\n(iii) \\(\\frac{3}{4}\\)
\n(iv) \\(\\frac{4}{3}\\)
\nAnswer:
\n(iv) \\(\\frac{4}{3}\\)<\/p>\n

Question 35.
\nThe value of sec \u03b8 cosec \u03b8 tan \u03b8 is :
\n(i) sec2<\/sup>\u03b8
\n(ii) cosec2<\/sup>\u03b8
\n(iii) cos2<\/sup>\u03b8
\n(iv) cos\u03b8.
\nAnswer:
\n(i) sec2<\/sup>\u03b8<\/p>\n

Question 36.
\nThe value of cos 240\u00b0 is :
\n(i) –\\(\\frac{\\sqrt{3}}{2}\\)
\n(ii) –\\(\\frac{1}{2}\\)
\n(iii) \\(\\frac{1}{2}\\)
\n(iv) \\(\\frac{\\sqrt{3}}{2}\\)
\nAnswer:
\n(ii) –\\(\\frac{1}{2}\\)<\/p>\n

Question 37.
\nThe value of \\(\\frac{2 \\tan 15^{\\circ}}{1+\\tan ^{2} 15^{\\circ}}\\) is :
\n(i) \\(\\frac{\\sqrt{3}}{2}\\)
\n(ii) \\(\\frac{1}{\\sqrt{2}}\\)
\n(iii) \\(\\frac{1}{2}\\)
\n(iv) \\(\\frac{1}{\\sqrt{3}}\\)
\nAnswer:
\n(iii) \\(\\frac{1}{2}\\)<\/p>\n

Question 38.
\nThe value of \"Introduction\u00a0is :
\n(i) \\(\\frac{1}{2}\\)
\n(ii) \\(\\frac{\\sqrt{3}}{2}\\)
\n(iii) \\(\\frac{2}{\\sqrt{3}}\\)
\n(iv) \\(\\frac{\\sqrt{2}}{1}\\)
\nAnswer:
\n(iii) \\(\\frac{2}{\\sqrt{3}}\\)<\/p>\n

Question 39.
\nThe value of sin (-300\u00b0) will be :
\n(i) \\(\\frac{\\sqrt{3}}{2}\\)
\n(ii) –\\(\\frac{\\sqrt{3}}{2}\\)
\n(iii) \\(\\frac{1}{2}\\)
\n(iv) –\\(\\frac{1}{2}\\)
\nAnswer:
\n(i) \\(\\frac{\\sqrt{3}}{2}\\)<\/p>\n

Question 40.
\nThe value of sin (-1080\u00b0) will be:
\n(i) -1
\n(ii) 0
\n(iii) 1
\n(iv) \u221e
\nAnswer:
\n(ii) 0<\/p>\n

\"Introduction<\/p>\n

Question 41.
\nThe measure of the angle subtended at the centre of a circle of radius 10 min radian by the arc of measure 5\u03c0 m will be :
\n(i) \\(\\frac{\\pi}{2}\\)
\n(ii) \\(\\frac{\\pi}{4}\\)
\n(iii) \\(\\frac{\\pi}{5}\\)
\n(iv) \\(\\frac{\\pi}{10}\\)
\nAnswer:
\n(i) \\(\\frac{\\pi}{2}\\)<\/p>\n

Question 42.
\nIf tan \u03b1 = sin \u03b1, then the value of \u03b1 will be:
\n(i) 90\u00b0
\n(ii) 60\u00b0
\n(iii) 45\u00b0
\n(iv) 0\u00b0
\nAnswer:
\n(iv) 0\u00b0<\/p>\n

Question 43.
\nThe value of 9 sec2<\/sup> \u03b8 – 9 tan2<\/sup> \u03b8 is :
\n(i) 1
\n(ii) 8
\n(iii) 9
\n(iv) 10
\nAnswer:
\n(iii) 9
\nSolution.
\n9 sec2<\/sup> \u03b8 – 9 tan2<\/sup> \u03b8 = 9 ( sec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 ) = 9 \u00d7 1 = 9
\nHence, choice (iii) is correct<\/p>\n

Question 44.
\nIt tan \u03b8 = \\(\\frac{a}{b}\\), then the value of \\(\\frac{b \\sin \\theta-a \\cos \\theta}{b \\sin \\theta+a \\cos \\theta}\\) will be :
\n(i) 1
\n(ii) \\(\\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\\)
\n(iii) \\(\\frac{b^{2}-a^{2}}{b^{2}+a^{2}}\\)
\n(iv) 0
\nAnswer:
\n(iv) 0<\/p>\n

\"Introduction<\/p>\n

Question 45.
\nThe value of \\(\\frac{2 \\tan 30^{\\circ}}{1-\\tan ^{2} 30}\\) will be
\n(i) cos 60\u00b0
\n(ii) sin 60\u00b0
\n(iii) tan 60\u00b0
\n(iv) cot 60\u00b0
\nAnswer:
\n(iii) tan 60\u00b0
\nSolution.
\n\"Introduction
\n= \u221a3 = tan 60\u00b0
\nHence choice (iii) is correct. (iii)<\/p>\n

Introduction to Trigonometry Class 10 Extra Questions Very Short Answer Type<\/h3>\n

Question 1.
\nFind the value of tan \\(\\left(\\frac{11 \\pi}{6}\\right)\\) :
\nSolution :
\ntan \\(\\left(\\frac{11 \\times 180}{6}\\right)\\)
\n= tan (11 \u00d7 30)
\n= tan 330\u00b0
\n= tan (360\u00b0 – 30\u00b0)
\n= tan 30\u00b0
\n= – \\(\\frac{1}{\\sqrt{3}}\\)<\/p>\n

Question 2.
\nIf 3x1<\/sub> = cosec \u03b8 and \\(\\frac{3}{x_{2}}\\) = cot \u03b8 find the value of \"Introduction
\nSolution.
\nGiven, 3x1<\/sub> = cosec \u03b8
\n\u2234 x1<\/sub> = \\(\\frac{1}{3}\\) cosec \u03b8
\n\"Introduction<\/p>\n

\"Introduction<\/p>\n

Question 3.
\nProve that
\n\"Introduction
\nSolution.
\n\"Introduction<\/p>\n

Question 4.
\nFind the value of \\(\\frac{\\sin 27^{\\circ}}{\\cos 63^{\\circ}}\\).
\nSolution.
\n\"Introduction<\/p>\n

Question 5.
\nIf cos A = \\(\\frac{\\sqrt{3}}{2}\\) , then find the value of sin 2A.
\nSolution.
\nGiven, cos A = \\(\\frac{\\sqrt{3}}{2}\\) = cos 30\u00b0
\n\u2234 A = 30\u00b0
\nsin 2A = sin 2 \u00d7 30\u00b0 = sin 60\u00b0 = \\(\\frac{\\sqrt{3}}{2}\\)<\/p>\n

Question 6.
\nIf tan 2A = cot (A – 18\u00b0), where 2A is an acute angle find the value of A.
\nSolution.
\ntan 2A = cot (A – 18\u00b0)
\n\u21d2 cot (90\u00b0 – 2A) = cot (A – 18\u00b0) [as cot (90 – \u03b8) = tan \u03b8]
\n\u21d2 90\u00b0 – 2A = A – 18\u00b0
\n\u21d2 3A = 90\u00b0 + 18\u00b0 = 108\u00b0
\n\u2234 A = \\(\\frac{108}{3}\\) = 36\u00b0<\/p>\n

\"Introduction<\/p>\n

Question 7.
\nSolve the equation
\n\"Introduction
\nSolution.
\n\"Introduction
\n\u2234 \u03b8 = 60\u00b0<\/p>\n

Question 8.
\nProve that sin4<\/sup> \u03b8 – cos4<\/sup> \u03b8 = 2 sin2<\/sup> \u03b8 – 1.
\nSolution.
\nL.H.S. = sin4<\/sup> \u03b8 – cos4<\/sup> \u03b8
\n= (sin2<\/sup> \u03b8)2<\/sup> \u2013 (cos2<\/sup> \u03b8)2<\/sup>
\n= (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) (sin2<\/sup> – cos2<\/sup> \u03b8)
\n= 1[sin2<\/sup> – (1 – sin a )]
\n= 2 sin2<\/sup> \u03b8 – 1 = R.H.S.<\/p>\n

Question 9.
\nIf cosec A = 2, Find the value of \"Introduction
\nSolution.
\nGiven, cosec A = 2 = cosec 30\u00b0
\n\u2234 A = 30\u00b0 sin A
\n\"Introduction
\n\"Introduction
\n= \u221a3 + 2 – \u221a3
\n= 2<\/p>\n

Question 10.
\nIn \u2206ABC, Prove :
\n\"Introduction
\nSolution.
\nIn \u2206ABC
\nA + B + C = 180\u00b0(angle sum property of \u2206)
\n= A + B = 180\u00b0 – C
\n\"Introduction
\n[\u2234 cos (90\u00b0 – \u03b8) = sin \u03b8]<\/p>\n

Question 11.
\nIn a right \u2206ABC, right angled at point C. If tan A = 1 prove that 2 sin A cos A = 1.
\n\"Introduction
\nSolution.
\ntan A = 1 = \\(\\frac{BC}{CA}\\)
\n\u2234 CA = BC = x (let)
\nBy pythagoras theorem
\nBA2<\/sup> = BC2<\/sup> + CA2<\/sup>
\n= x2<\/sup> + x2<\/sup> = 2x2<\/sup>
\n\u2234 BA = x \u221a2
\n\"Introduction
\n\u2234 L.H.S. = 2. sin A cos A
\n= 2.\\(\\frac{1}{\\sqrt{2}}\\) \u00d7 \\(\\frac{1}{\\sqrt{2}}\\)
\n= 1 = R.H.S<\/p>\n

\"Introduction<\/p>\n

Question 12.
\nProve
\n\"Introduction
\n\"Introduction
\nSolution.
\n(i) R.H.S. = (sec \u03b8 – tan \u03b8)2<\/sup>
\n\"Introduction<\/p>\n

(ii) R.H.S. (cosec \u03b8 + cot \u03b8)2<\/sup>
\n\"Introduction<\/p>\n

Question 13.
\nProve (1 – sin \u03b8) (1 + sin \u03b8) (1 + tan2<\/sup> \u03b8) = 1
\nSolution.
\nL.H.S. = (1 – sin \u03b8) (1 + sin \u03b8) (1 + tan2<\/sup> \u03b8)
\n= (1 – sin2<\/sup> \u03b8) (sec2<\/sup> \u03b8)
\n[\u2235 1 – sin2<\/sup> = cos2<\/sup> \u03b8, 1 + tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8]
\n= cos2<\/sup> \u03b8 \u00d7 sec2<\/sup> \u03b8
\n= cos2<\/sup>\u03b8 \u00d7 \\(\\frac{1}{\\cos ^{2} \\theta}\\)
\n= 1 = R.H.S.<\/p>\n

Question 14.
\nProve : (1 + cot \u03b8 + tan \u03b8) (sin \u03b8 – cos \u03b8)
\n= \"Introduction
\nSolution.
\nL.H.S. = (1 + cot \u03b8 +, tan \u03b8) (sin \u03b8 – cos \u03b8)
\n\"Introduction<\/p>\n

Question 15.
\nProve that:
\ncot A – cot 2A = cosec 2A.
\nSolution :
\nL.H.S. = cot A – cot 2A
\n\"Introduction
\n= cosec 2A = R.H.S.<\/p>\n

Question 16.
\nFind the value of :
\n(i) Find the value of top \\(\\frac{13 \\pi}{6}\\)
\n(ii) Find the value of sec \\(\\frac{23 \\pi}{4}\\)
\n(iii) Find the value of cot \\(\\frac{13 \\pi}{3}\\)
\n(iv) Find the value of (cos 75\u00b0 + cos 15\u00b0).
\nSolution.
\n\"Introduction<\/p>\n

Question 17.
\nProve that :
\n\"Introduction
\nSolution :
\nL. H. S.
\n\"Introduction
\n= R.H.S.<\/p>\n

Question 18.
\nIf cos (-840\u00b0) = –\\(\\frac{1}{2}\\) then find the value of sin (-840\u00b0).
\nSolution.
\ncos (- 840\u00b0) = cos (840\u00b0)
\n= cos (120\u00b0) = –\\(\\frac{1}{2}\\)
\nHence
\nsin (-840\u00b0) = – sin 120\u00b0 = \\(\\frac{-\\sqrt{3}}{2}\\)<\/p>\n

\"Introduction<\/p>\n

Question 19.
\nIf sin \u03b8 = \\(\\frac{4}{5}\\), then find the value of cos 2\u03b8.
\nSolution.
\nsin \u03b8 = \\(\\frac{4}{5}\\)
\ncos 2\u03b8 = 1 – 2 sin2<\/sup>\u03b8
\n= 1 – 2(\\(\\frac{4}{5}\\))2<\/sup> = –\\(\\frac{7}{25}\\)<\/p>\n

Question 20.
\nFind the value of tan 35\u00b0 tan 40\u00b0 tan 45\u00b0 tan 50\u00b0 tan 55\u00b0
\nSolution.
\ntan 35o tan 40\u00b0 tan 45o tan 50\u00b0 tan 55\u00b0
\n= tan (90\u00b0 – 55\u00b0) tan (90\u00b0 – 50\u00b0) .1.tan 50\u00b0 tan 55\u00b0
\n= cot 55\u00b0 cot 50\u00b0.1.tan 50\u00b0 tan 55\u00b0
\n= 1.<\/p>\n

Question 21.
\nProve that :
\n\"Introduction
\nSolution :
\nR. H. S.
\n\"Introduction<\/p>\n

Question 22.
\nProve.
\n\"Introduction
\nSolution.
\nL. H. S.
\n\"Introduction
\n\"Introduction<\/p>\n

Question 23.
\nProve that : (cos A + cos B)2<\/sup> + (sin A + sin B)2<\/sup>
\n\"Introduction
\nSolution :
\nL.H.S.
\ncos2<\/sup> A + cos2<\/sup> B + 2 cos A cos B
\n+ sin2<\/sup> A + sin2<\/sup> B + 2 sinA sinB
\n\u21d2 1 + 1 + 2 (cos A cos B + sin A sin B)
\n\u21d2 2 + 2 cos (A – B)
\n\u21d2 2 [1 + cos (A – B)]
\n\"Introduction<\/p>\n

\"Introduction<\/p>\n

Question 24.
\nProve the following:
\n(i) \"Introduction
\n(ii) \"Introduction
\n(iii) \"Introduction
\n(iv) Find the value of
\n\"Introduction
\n(v) Find the value of
\n\"Introduction
\n(vi) Find the value of sin 3A in terms of sin A. Hence find the value of sin 135\u00b0. If A = 45\u00b0.
\n(vii) If sin \u03b8 = \\(\\frac{3}{25}\\), find the value of sin 2\u03b8
\nSolution.
\n(i)
\n\"Introduction
\n(ii)
\n\"Introduction
\n(iii)
\n\"Introduction
\n(iv)
\n\"Introduction
\n(v)
\n\"Introduction
\n(vi) sin 3A = sin(2A + A)
\n= sin 2A. cos A + cos 2A.sin A
\n= 2sin A.cos A. cos A + (1 – 2 sin2<\/sup>A).sin A
\n= 2 sin A(1 – sin2<\/sup>A) + sin A – 2 sin3<\/sup>A
\n= 2 sin A – 2 sin3<\/sup>A + sin A – 2 sin3<\/sup>A
\n= 3 sin A – 4 sin3<\/sup>A.
\n\u2234 sin 135\u00b0 = sin(3 \u00d7 45\u00b0)
\n= 3 sin 45\u00b0 – 4 sin3<\/sup> 45\u00b0
\n\"Introduction<\/p>\n

(vii) As sin 2 \u03b8 = 2 sin \u03b8 . cos \u03b8
\n\"Introduction<\/p>\n

Question 25.
\nProve that : sec \u03b8 (1 – sin \u03b8) (sec \u03b8 + tan \u03b8) = 1
\nSolution :
\nL.H.S.
\n\"Introduction<\/p>\n

Question 26.
\nProve that :
\ncos 33\u00b0cos 27 – cos 57\u00b0cos 63\u00b0 = \\(\\frac{1}{2}\\)
\nSolution :
\nL.H.S.
\ncos 33\u00b0 cos 27\u00b0 – cos (90\u00b0-33\u00b0) cos (90\u00b0 – 27\u00b0)
\n\u2234 cos 33\u00b0 cos 27\u00b0 – sin 33\u00b0 sin27\u00b0
\n\u2234 cos (33\u00b0 + 27\u00b0) = cos (60\u00b0)
\n= \\(\\frac{1}{2}\\)<\/p>\n

\"Introduction<\/p>\n

Question 27.
\nFind the value of :
\ncos 80o.cos 70\u00b0 – cos 10\u00b0 cos 20\u00b0.
\nSolution.
\ncos 80\u00b0 cos 70\u00b0 – cos 10\u00b0 cos 20\u00b0
\n= cos(90\u00b0 – 10\u00b0) cos(90\u00b0 – 20\u00b0) – cos 10\u00b0 cos 20\u00b0
\n= sin 10\u00b0 sin 20\u00b0 – cos 10\u00b0 cos 20\u00b0
\n= – (cos 10\u00b0 cos 20\u00b0 – sin 10\u00b0 sin 20\u00b0)
\n= -cos(10\u00b0 + 20\u00b0)
\n= -cos 30\u00b0 = –\\(\\frac{-\\sqrt{3}}{2}\\).<\/p>\n

Question 28.
\nProve :
\n\"Introduction
\nSolution.
\n\"Introduction
\nL.H.S. sin 5A – 2 sin 3A + sin A cos 5A – 2 cos 3A + cos A
\nsin 5A +sin A – 2 sin 3A , cos 5A + cos A \u2013 2 cos 3A
\n2sin 3A cos 2A \u2013 2sin 3A 2 cos 3A cos 2A – 2 cos 3A 2sin 3A(cos 2A – 1)
\n2cos 3A (cos 2A – 1) = tan 3A = R.H.S.<\/p>\n

Question 29.
\nProve cos2<\/sup> \u03c0\/8 + cos2<\/sup>3\u03c0 + cos2<\/sup> 5\u03c0\/8 + cos2<\/sup> 7\u03c0\/8 = 2.
\nSolution :
\n\"Introduction
\n1 + 1 = 2 = R.H.S.<\/p>\n

Question 30.
\nProve :
\ncos 4A = 1 – 8 sin2<\/sup>A + 8 sin4<\/sup>A
\nSolution :
\nL. H. S.
\ncos 4A = cos 2(2A)
\n= 1 – 2 sin\u03b8 2A
\n= 1 – 2 (2 sin A cosA)2<\/sup>
\n= 1 – 8 sin2<\/sup> A cos2<\/sup>A
\n= 1 – 8 sin – A (1 – sin2<\/sup> A)
\n= 1 – 8 sin2<\/sup> A + 8 sin4<\/sup>A
\n= R.H.S.<\/p>\n

Introduction to Trigonometry Class 10 Extra Questions Short Answer Type<\/h3>\n

Question 1.
\nProve that :
\n\"Introduction
\nSolution :
\nL.H.S.
\n\"Introduction
\n= 1 + 1 = 2 = R.H.S.<\/p>\n

Question 2.
\nProve that:
\n\"Introduction
\nSolution :
\nL. H. S.
\n\"Introduction
\n= 2 \u00d7 cosec A = R.H.S.<\/p>\n

\"Introduction<\/p>\n

Question 3.
\nProve that :
\n\\(\\frac{\\sin 8 A}{\\sin A}\\) = 8 cos A cos 2A cos 4A
\nSolution :
\n\"Introduction
\n= 8 cos A cos 2A cos 4A = R.H.S<\/p>\n

Question 4.
\nProve that : (1 – sin \u03b8)(1 + sin \u03b8)( 1+ tan2\u03b8)= 1
\nSolution :
\nL.H.S.
\n= (1- sin A)(1 + sin )( 1+ tan2\u03b8)
\n= (1 – sin2<\/sup> \u03b8) \u00d7 sec2<\/sup>\u03b8
\n= cos2<\/sup>\u03b8 \u00d7 \\(\\frac{1}{\\cos ^{2} \\theta}\\) = 1
\n= R.H.S.<\/p>\n

Question 5.
\nProve that : sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A
\nSolution.
\n\"Introduction
\n= sec A + cosec A R.H.S.<\/p>\n

Question 6.
\nFind the value of
\n\"Introduction
\nSolution :
\n(cos 15\u00b0 + sin 15\u00b0)
\n\"Introduction<\/p>\n

Question 7.
\nProve that : 16 cos 20\u00b0 cos 40\u00b0 cos 60\u00b0 cos 80\u00b0 = 1
\nSolution:
\nL.H.S.= 16 cos 20\u00b0 cos 40\u00b0 cos 60\u00b0 cos 80\u00b0
\n= 8 \u00d7 \\(\\frac{1}{2}\\) (2 cos 80\u00b0 cos 40\u00b0) cos 20\u00b0
\n= 4 (cos 120\u00b0 cos 40\u00b0) cos 200 = 2. 2. – cos 20\u00b0 + cos 40ocos 2007
\n= 2 [-cos 20\u00b0 + cos 60\u00b0 + cos 20\u00b0]
\n= 2 \u00d7 \\(\\frac{1}{2}\\) = 1 = R.H.S.<\/p>\n

Question 8.
\nProve that:
\n\"Introduction
\nSolution :
\nL.H.S.
\n= sec2<\/sup>A – sec2<\/sup>A – tan2<\/sup>A \u00d7 tan2<\/sup>A
\n= sec4<\/sup> A – tan4<\/sup>A
\n= (sec2<\/sup>A -tan2<\/sup>A)(sec2<\/sup>A + tan2<\/sup>A)
\n= 1 (1 + tan2<\/sup>A + tan2<\/sup>A)
\n= 1 + 2 tan2A = R.H.S.<\/p>\n

Question 9.
\nProve that :
\n\"Introduction
\nSolution :
\nL.H.S. =
\n\"Introduction
\n= 2 cot \u03b8 = R.H.S.<\/p>\n

\"Introduction<\/p>\n

Question 10.
\nProve that :
\n(i) If 2 tan P= 3 tan Q, then prove that:
\ntan(P – Q) = \\(\\frac{\\sin 2 Q}{5-\\cos 2 Q}\\)
\n(ii) If cos A + sin A = \u221a2 cos A, prove that cos A – sin A = \u221a2 sin A . sin 63\u00b0 + cos 630
\n\"Introduction
\nSolution.
\n\"Introduction<\/p>\n

(ii) Given cos A + sin A = \u221a2cos A
\n\u2234 sin A = (\u221a2 – 1) cos A Now, multiply both sides by (\u221a2 +1)
\n\u2234 (\u221a2 + 1)sin A = (\u221a2 + 1) (\u221a2 – 1) cosA
\n= \u221a2 sin A + sin A = cos A
\nHence, \u221a2sin A = cos A – sin A.<\/p>\n

(iii)
\n\"Introduction
\n\"Introduction<\/p>\n

(iv) L.H.S.
\n\"Introduction<\/p>\n

Question 11.
\nIf cosec A = \\(\\frac{17}{15}\\), then find the value of sec A.
\nSolution.
\ncosec A = \\(\\frac{17}{15}\\)
\nsin A = \\(\\frac{15}{17}\\)
\ncos A = \\(\\frac{8}{17}\\)
\n\u2235 sec A = \\(\\frac{17}{8}\\)<\/p>\n

Question 12.
\nProve that
\n\"Introduction
\n(i) \u03b8, (ii) cos2<\/sup> \u03b8 + cos2<\/sup> \u03b8.cot2<\/sup> \u03b8 = cot2<\/sup> \u03b8
\nSolution.
\n\"Introduction
\n(ii) L.H.S. = cos2<\/sup> \u03b8 + cos2<\/sup> x cot2<\/sup> \u03b8
\n= cos2<\/sup> \u03b8[1 + cot2<\/sup> \u03b8]
\n= cos2<\/sup> \u03b8 \u00d7 cosec2<\/sup>\u03b8
\n= cos \u03b8 \u00d7 \\(\\frac{1}{\\sin ^{2} \\theta}\\) = cot2<\/sup> \u03b8 = R.H.S.<\/p>\n

Question 13.
\nSolve the equation :
\n\"Introduction
\nSolution.
\n\"Introduction<\/p>\n

\"Introduction<\/p>\n

Question 14.
\nIf (A + B) = 45\u00b0, then prove that (1 + tan A) (1 + tan B) = 2
\nSolution.
\nA + B = 45\u00b0
\n\u21d2 tan(A + B) = tan 45o
\n\u21d2 \\(\\frac{\\tan A+\\tan B}{1-\\tan A \\tan B}\\) = 1
\n\u21d2 tan A + tan B = 1 – tan A tan B
\n\u21d2 tan A + tan B + tan A tan B = 1
\n\u21d2 1 + tan A + tan B + tan A tan B = 1 + 1
\n(1 + tan A) + tan B(1 + tan A) = 2
\n(1 + tan A)(1+tan B) = 2<\/p>\n

Question 15.
\nProve that tan 75\u00b0 = 2 + \u221a3.
\nSolution.
\ntan 75\u00b0
\n= tan (45\u00b0 + 30\u00b0)
\n\"Introduction
\n\"Introduction<\/p>\n

Question 16.
\nProve that
\n\"Introduction
\nSolution.
\n\"Introduction<\/p>\n

Question 17.
\nFind the value of \\(\\frac{\\sin 65^{\\circ}}{\\sin 115^{\\circ}}\\)
\nSolution.
\nWe have,
\n\"Introduction<\/p>\n

Question 18.
\nProve that
\n\"Introduction
\nSolution.
\n\"Introduction<\/p>\n

Question 19.
\nProve that sec 70\u00b0 sin 20\u00b0 + cos 20\u00b0 cosec 70\u00b0 = 2
\nSolution.
\nL. H. S.
\nsec 70\u00b0 sin 20\u00b0 + cos 20\u00b0 cosec 70\u00b0
\n\"Introduction<\/p>\n

Question 20.
\nProve that :
\n(i) cos 2A = \\(\\frac{1-\\tan ^{2} A}{1+\\tan ^{2} A}\\)
\n(ii) \"Introduction
\nSolution.
\n(i) R.H.S
\n\"Introduction<\/p>\n

(ii)
\n\"Introduction<\/p>\n

\"Introduction<\/p>\n

Question 21.
\nIf sec \u03b8 + tan \u03b8 = p, prove that
\n\\(\\frac{p^{2}-1}{p^{2}+1}\\) = sin \u03b8.
\nSolution.
\n\"Introduction
\n\"Introduction<\/p>\n

Question 22.
\nIf A = 30\u00b0 and B = 60\u00b0, then prove that sin(A + B) = sin A cos B + cos A sin B.
\nSolution.
\nL.H.S. = sin (A + B)
\n= sin (30\u00b0 + 60\u00b0)
\n= sin 90\u00b0 = 1
\nR.H.S. = sin A cos B + cos A sin B
\n= sin 30\u00b0. cos 60\u00b0 + cos 30\u00b0.sin 60\u00b0
\n\"Introduction
\nL.H.S. = R.H.S.<\/p>\n

Question 23.
\nFind the value of sin 65o cos 35\u00b0 – cos 65\u00b0 sin 35\u00b0.
\nSolution.
\nsin 65\u00b0 cos 35\u00b0 – cos 65\u00b0 sin 35\u00b0
\n= sin(65\u00b0 – 35\u00b0)
\n= sin 30\u00b0 = \\(\\frac{1}{2}\\)<\/p>\n

Question 24.
\nIf sin\u03b8 = \\(\\frac{9}{41}\\), then find the value of tan \u03b8 – cosec \u03b8.
\nSolution.
\n\"Introduction<\/p>\n

Question 25.
\nFind the value of sin A cos(90\u00b0 – A) + cos A sin(90\u00b0 – A).
\nSolution.
\nsin A cos (90\u00b0 – A) + cos A sin (90\u00b0 – A)
\n= sin A sin A + cos A cos A
\n= sin2<\/sup>A + cos2<\/sup>A = 1<\/p>\n

Question 26.
\nProve that :
\n\\(\\sqrt{\\frac{1-\\cos \\theta}{1+\\cos \\theta}}\\) = cosec \u03b8 – cot \u03b8
\nSolution.
\n\"Introduction
\ncosec \u03b8 – cot \u03b8 = R.H.S.<\/p>\n

\"Introduction<\/p>\n

Question 27.
\n(a) Prove that :
\n\"Introduction
\nOr
\nsin 4\u03b8 – cos 4\u03b8 = 2 tan2<\/sup>\u03b8 – 1.
\n(b) Prove that :
\ncos4<\/sup>\u03b8 – sin2<\/sup>\u03b8 = cos 2\u03b8.
\n(c) Find the value of cos 375\u00b0.
\nSolution.
\n(a) Do yourself.
\n(b) Do yourself.
\n(c) cos 375\u00b0 = cos (360\u00b0 + 15\u00b0)
\n= cos 15o = cos (45\u00b0 \u2013 30\u00b0)
\n= cos 45\u00b0.cos 30\u00b0 + sin 45o.sin 30\u00b0
\n\"Introduction<\/p>\n

Question 28.
\nProve that: sin 3A cos3A + cos 3A. sinA = \\(\\frac{3}{4}\\) sin 4A.
\nSolution.
\nDo yourself.<\/p>\n

Question 29.
\nProve that :
\n\"Introduction
\nSolution.
\nDo yourself.<\/p>\n

Question 30.
\n(a) Prove that:
\n\"Introduction
\nSolution.
\n\"Introduction<\/p>\n

Question 31.
\nProve :
\n\\(\\frac{\\sec A-\\tan A}{\\sec A+\\tan A}\\)
\n= 1 – 2 sec A tan A + 2 tan2<\/sup>A
\nSolution.
\nL.H.S. – \\(\\frac{\\sec A-\\tan A}{\\sec A+\\tan A}\\)
\n\"Introduction
\n= 1 + tan2<\/sup>A + tan2<\/sup> – A \u2013 2sec A tan A
\n= 1 – 2 sec A. tan A + 2 tan2<\/sup>A
\n= R.H.S.<\/p>\n

\"Introduction<\/p>\n

Question 32.
\nProve :
\n2cos A + cos 3A + cos 5A
\n= 4 cos A cos2<\/sup> 2A.
\nSolution.
\nL.H.S. = 2 cos A + cos 3A + cos 5A
\n= 2 cos A + 2 cos 4A cos (-A)
\n= 2 cos A + 2 cos 4A cos A
\n= 2 cos A(1 + cos 4A)
\n= 2 cos A(1 + 2 cos2<\/sup>2A – 1)
\n= 4 cos A cos2<\/sup>2A = R.H.S.<\/p>\n

Question 33.
\nProve : sin2<\/sup>A – sin2<\/sup>B = sin(A + B) sin(A – B).
\nSolution.
\nL.H.S. = sin2<\/sup>A – sin2<\/sup>B
\n= (sin A – sin B) (sin A + sin B)
\n\"Introduction
\n= sin(A + B).sin(A – B)
\n= R.H.S.<\/p>\n

Question 34.
\nProve :
\n\"Introduction
\nSolution.
\nL.H.S.
\n\"Introduction
\n\"Introduction
\nL.H.S. = R.H.S.<\/p>\n

Question 35.
\nProve :
\nProved.
\nsec A – tan A = \\(\\frac{1}{\\sec A+\\tan A}\\)
\nSolution.
\nL.H.S. = sec A \u2013 tan A
\n\"Introduction
\nSo, L.H.S. = R.H.S.<\/p>\n

Question 36.
\nProve that :
\n\"Introduction
\nSolution.
\nL.H.S.
\n\"Introduction
\n\"Introduction<\/p>\n

Introduction to Trigonometry Class 10 Extra Questions Long Answer Type<\/h3>\n

Question 1.
\nProve that:
\n(cos A + cos B)2<\/sup> + (sin A – sin B)2<\/sup>
\n= 4 cos2<\/sup> \\(\\left(\\frac{A+B}{2}\\right)\\)
\nSolution.
\nL.H.S.
\ncos2<\/sup>A + cos2<\/sup>B + 2cos A cos B + sin2<\/sup>A
\n+ sin2<\/sup>B – 2 sin A sinB
\n\u21d2 (cosA + sinoA) + (cos’B + sino B)
\n+ 2 (cosA cos B – sin A sin B)
\n\u21d2 1 + 1 + 2 cos (A – B)
\n\u21d2 2[1 + cos (A – B)]
\n\"Introduction
\n= 4cos2<\/sup> \\(\\frac{A-B}{2}\\)
\n= R.H.S.<\/p>\n

\"Introduction<\/p>\n

Question 2.
\nProve that:
\n\\(\\sqrt{\\frac{1-\\sin A}{1+\\sin A}}\\)= sec A – tan A
\nSolution.
\n\"Introduction
\n= sec A – tan A
\n= R.H.S.<\/p>\n

Question 3.
\nProve that :
\n\"Introduction
\nSolution :
\nL.H.S.
\n\"Introduction
\n\"Introduction
\n= sec A- sec A + tan A
\n= tan A ……(ii)
\nFrom (i) and (ii)
\nL.H.S. = R.H.S.<\/p>\n

Question 4.
\nProve that : cos3<\/sup>A cos3A + sin3<\/sup>A sin3A = cos3<\/sup>2A.
\nSolution :
\n\"Introduction
\n= cos3<\/sup> 2A = R.H.S.<\/p>\n

Question 5.
\nIf \\(\\frac{\\cos \\alpha}{\\cos \\beta}\\) = n, \\(\\frac{\\sin \\alpha}{\\sin \\beta}\\) = m prove that
\n(m2<\/sup> – n2<\/sup>) sin2<\/sup>\u03b2 = 1 – n2<\/sup>.
\nSolution.
\ncos \u03b1 = n cos \u03b2 and sin \u03b1 = m sin \u03b2
\nSquaring and adding we get
\ncos2<\/sup>\u03b1 + sin2<\/sup> \u03b1 = n2<\/sup> cos2<\/sup>\u03b2 + m2<\/sup> sin2<\/sup> \u03b2.
\n\u21d2 1 = n2<\/sup> (1 – sin2<\/sup>\u03b2) + m2<\/sup> sin2<\/sup> \u03b2
\n\u21d2 1 = n2<\/sup> – n2<\/sup> sin2<\/sup> \u03b2 + m2 sino \u03b2
\n\u21d2 1 – n2<\/sup> = sin2<\/sup>\u03b2 (m2<\/sup> – n2<\/sup>)<\/p>\n

Question 6.
\nProve that :
\nsin 20\u00b0 sin 40\u00b0 sin 80\u00b0 = \\(\\sqrt{\\frac{3}{8}}\\)
\nSolution :
\n\"Introduction<\/p>\n

Question 7.
\nProve :
\n\"Introduction
\nSolution :
\n\"Introduction
\n\"Introduction<\/p>\n

Question 8.
\nProve :
\n\"Introduction
\nSolution.
\nL.H.S.
\n\"Introduction
\n= (cosec \u03b8 + cot \u03b8)2<\/sup> = R.H.S. Proved<\/p>\n

Question 9.
\nProve:
\ntan2<\/sup> \u03b8 + cot2<\/sup> \u03b8 + 2 = sec2<\/sup> \u03b8.cosec2<\/sup> \u03b8
\nSolution.
\ntan2<\/sup>\u03b8 + cot2<\/sup> \u03b8 + 2
\n= (1 + tan2<\/sup>\u03b8) + (1 + cot2<\/sup>\u03b8)
\n= sec2<\/sup> \u03b8 + cosec2<\/sup> \u03b8
\n\"Introduction
\n= sec2<\/sup> \u03b8 cosec2<\/sup> \u03b8 = R.H.S.<\/p>\n

\"Introduction<\/p>\n

Question 10.
\nIf A + B + C = 180\u00b0, Prove that :
\n(i) sin A + sin B + sin C = 4 cos\\(\\frac{A}{2}\\) . cos\\(\\frac{B}{2}\\) . cos\\(\\frac{C}{2}\\)
\n(ii) cot A . cot B + cot B . cot C + cot C . cot A = 1
\n(iii) sin 2A + sin 2B + sin 2C = 4 sin A sin B . sin C
\n(iv) cos A + cos B + cos C = 1 + 4 sin\\(\\frac{A}{2}\\) . sin\\(\\frac{B}{2}\\) . sin\\(\\frac{C}{2}\\)
\n(v) sin A + sin B – sin C = 4 sin\\(\\frac{A}{2}\\) . sin\\(\\frac{B}{2}\\) . cos\\(\\frac{C}{2}\\)
\n(vi) Prove :
\n\"Introduction
\nSolution.
\n(i) L.H.S.
\n= sin A + sin B + sin C
\n\"Introduction<\/p>\n

(ii) We have,
\nA + B + C = 180\u00b0
\n\u21d2 A + B = 180\u00b0- C
\n\u2234 cot(A + B) = cot(180\u00b0 – C)
\n\u21d2 \\(\\frac{\\cot A \\cdot \\cot B-1}{\\cot B+\\cot A}\\) = -cot C
\n\u21d2 cot A . cot B – 1 = -cot B . cot C – cot A . cot C
\n\u21d2 cot A . cot B + cot B . cot C + cot C . cot A = 1
\n= R.H.S.<\/p>\n

(iii) L.H.S. = sin 2A + sin 2B + sin 2C
\n\"Introduction
\n= 2 sin(A + B) . cos(A – B) + 2 sin C.cos C…(i)
\n\u2235 A+B+C = 180\u00b0
\n\u2234 A + B = 180\u00b0 – C
\n\u21d2 sin(A + B) = sin(180\u00b0 – C)
\n= sin C
\nand cos (A + B) = cos (180\u00b0 – C)
\n= -cos C
\nPutting value in (i),
\n= 2 sin C.cos(A – B) – 2 sin C. cos(A + B)
\n= 2 sin C[cos (A – B) – cos(A + B)]
\n\"Introduction
\n= 2 sin C(2 sin A.sin B]
\n= 4 sin A . sin B . sin C
\n= R.H.S.<\/p>\n

(iv) L.H.S. = cos A + cos B + cos C
\n\"Introduction
\n= R.H.S.<\/p>\n

(v) Do yourself.<\/p>\n

(vi) L.H.S.
\n\"Introduction
\n\"Introduction<\/p>\n

Question 11.
\nProve that:
\n\"Introduction
\n= cos 2\u03b8 – tan 3\u03b8 . sin 2\u03b8.
\n(ii) sin 10\u00b0 . sin 30\u00b0 . sin 50\u00b0 . sin 70\u00b0 = \\(\\frac{1}{16}\\)
\n(iii) cos 20\u00b0 . cos 40\u00b0 cos 60\u00b0. cos 80\u00b0 = \\(\\frac{1}{16}\\)
\n(iv) \"Introduction
\n(v) sin 20\u00b0.sin 40\u00b0 sin 60\u00b0.sin 80\u00b0 = \\(\\frac{3}{16}\\)
\n(vi) tan 70\u00b0 = tan 20\u00b0 + 2 tan 50\u00b0.
\n(vii) tan 3A – tan 2A – tan A = tan 3A . tan 2A tan A.
\n(viii) If A + B + C = 90\u00b0, prove that
\ntan A tan B + tan B tan C + tan C tan A = 1
\n(ix) Prove that :
\n\"Introduction
\n(x) Prove :
\n\"Introduction
\n+ (sin A + sin B)2<\/sup>
\nSolution.
\n(i) L.H.S.
\n\"Introduction<\/p>\n

= cos 2\u03b8 – tan 3\u03b8.sin 2\u03b8
\n= R.H.S.<\/p>\n

(ii) L.H.S. = sin 10\u00b0 sin 50\u00b0 sin 70\u00b0 sin 30\u00b0
\n\"Introduction<\/p>\n

(iii) L.H.S. = cos 20\u00b0 cos 40\u00b0.cos 80\u00b0.cos 60\u00b0
\n\"Introduction<\/p>\n

(iv)
\n\"Introduction<\/p>\n

(v) L.H.S.
\n= sin 20\u00b0 sin 40\u00b0 sin 60\u00b0 sin 80\u00b0
\n= sin 60\u00b0 sin 20\u00b0 sin 40\u00b0 sin 80\u00b0
\n\"Introduction
\nL.H.S. = R.H.S.<\/p>\n

(vi) tan 70\u00b0 = tan(50\u00b0 + 20\u00b0) .
\n\"Introduction
\n\u21d2 tan 70\u00b0 – tan 70\u00b0 tan 50\u00b0 tan 20\u00b0
\n= tan 50\u00b0 + tan 20\u00b0
\n\u21d2 tan 70\u00b0 – tan(90\u00b0 – 20\u00b0) tan 50\u00b0
\ntan 20\u00b0 = tan 50\u00b0 + tan 20\u00b0
\n\u21d2 tan 70\u00b0 – cot 20\u00b0 tan 50\u00b0 tan 20\u00b0
\n= tan 50\u00b0 + tan 20\u00b0
\n\u21d2 tan 70\u00b0 – \\(\\frac{1}{\\tan 20^{\\circ}}\\) tan 50\u00b0 tan 20\u00b0
\n= tan 50\u00b0 + tan 20\u00b0
\n\u21d2 tan 70\u00b0 – tan 50\u00b0 = tan 50\u00b0 + tan 20\u00b0
\n\u21d2 tan 70\u00b0 = tan 50\u00b0 + tan 20\u00b0+ tan 50\u00b0
\n\u21d2 tan 70\u00b0= tan 20\u00b0 + 2tan 50\u00b0<\/p>\n

(vii) We have
\ntan 3A = tan(A + 2A)
\ntan A + tan 2A
\n\u21d2 tan 3A = \\(\\frac{\\tan A+\\tan 2 A}{1-\\tan A \\tan 2 A}\\)
\n\u21d2 tan 3A – tan 3A tan A tan 2A = tan A + tan 2A
\n\u21d2 tan 3A – tan 2A – tan A
\n= tan 3A tan 2A tan A.<\/p>\n

(viii) Given
\nA + B + C = 90\u00b0
\n\u21d2 A + B = 90\u00b0 – C
\n\u21d2 tan(A + B) = tan(90\u00b0 – C)
\n\u21d2 \\(\\frac{\\tan A+\\tan B}{1-\\tan A \\tan B}\\) = cot C = \\(\\frac{1}{\\tan C}\\)
\n\u21d2 tan =tan A . tan C + tan B . tan C
\n= 1 – tan A tan B
\n\u21d2 tan A tan B + tan B . tan C + tan C . tan A = 1<\/p>\n

(ix)
\n\"Introduction
\n\"Introduction
\n= sec A + tan A
\n= R.H.S.<\/p>\n

(x) R.H.S. = (cos A-cos B)2<\/sup> + (sin A + sin B)2<\/sup>
\n= cos2<\/sup>A + cos2<\/sup>B – 2cos A cos B
\n+ sin2<\/sup>A + sin2<\/sup>B + 2 sin A sin B
\n= (cos2<\/sup> A +sin2<\/sup> A) + (cos2<\/sup>B + sin2<\/sup>B)
\n– 2{cos A cos B – sin A sin B]
\n= 1 + 1 – 2[cos A.cos B – sin A.sin B]
\n= 2[1 – cos(A + B)]
\n\"Introduction
\n= L.H.S.<\/p>\n

\"Introduction<\/p>\n

Question 12.
\nSolve :
\n\"Introduction
\nSolution.
\n\"Introduction
\n\u21d2 cos \u03b8 = cos 60\u00b0
\n\u2234 \u03b8 = 60\u00b0<\/p>\n

Question 13.
\nProve : (cos A + sec A)2<\/sup> + (sin A + cosec A)2<\/sup> 7 + tan2<\/sup> A + cot2<\/sup> A.
\nSolution.
\nL.H.S.
\n= (cos A + sec A)2<\/sup> + (sin A + cosec A)2<\/sup>
\n= cos2<\/sup>A + sec2<\/sup>A + 2.cos A sec A + sin2<\/sup> A + cosec2<\/sup>A + 2sin A cosec A
\n= 1 + 2 + 2 + sec2<\/sup>A + cosec2<\/sup> A
\n= 5 + 1 + tan2<\/sup>A + 1 + cot2<\/sup> A
\n= 7 + tan2<\/sup>A + cot2<\/sup>A
\n= R.H.S.<\/p>\n

Question 14.
\nProve :
\n\"Introduction
\nSolution :
\n\"Introduction<\/p>\n

Question 15.
\nProve that:
\ntan (A + B) tan (A – B)
\n\"Introduction
\nSolution.
\ntan (A + B)
\n\"Introduction<\/p>\n

Question 16.
\n\"Introduction
\nSolution.
\nL. H. S.
\n\"Introduction
\n\"Introduction
\n= tan 5 A = R.H.S.<\/p>\n

Question 17.
\nProve :
\ncos A cos (60\u00b0 + A). Cos (60\u00b0 – A) = \\(\\frac{1}{4}\\) cos 3A
\nSolution.
\nL.H.S.
\n= cos A. cos (60\u00b0 + A). cos (60\u00b0 – A)
\n= cos A (cos 60\u00b0. cos A-sin 60\u00b0 sin A)
\n(cos 60\u00b0. cos A + sin 60\u00b0sin A)
\n\"Introduction<\/p>\n

Question 18.
\nProve :
\n\"Introduction
\nSolution.
\nL.H.S.
\n\"Introduction
\n\"Introduction
\n= tan (A – B)
\n= R.H.S.<\/p>\n

\"Introduction<\/p>\n

Question 19.
\nProve :
\nsin A sin 2 A + sin 3A sin 6A = sin 4A sin 5A.
\nSolution.
\nL.H.S.
\n= sin A sin 2A + sin3 A sin 6A
\n= \\(\\frac{1}{2}\\) [2 sin A sin 2A + 2 sin3A sin 6A]
\n= \\(\\frac{1}{2}\\) [cos A – cos 3A + cos 3 A – cos 9A]
\n= \\(\\frac{1}{2}\\) [cos A – cos 9A]
\n= \\(\\frac{1}{2}\\) \u00d7 2 sin \\(\\frac{9 \\mathrm{A}+\\mathrm{A}}{2}\\) sin \\(\\frac{9 \\mathrm{A}-\\mathrm{A}}{2}\\)
\n= sin 5A sin 4A
\n= sin 4A sin 5A = R.H.S.<\/p>\n

Question 20.
\nProve :
\n\"Introduction
\nSolution.
\nL.H.S.
\n\"Introduction
\n= sec \u03b8 + tan \u03b8 = R.H.S.<\/p>\n

\"Introduction<\/p>\n","protected":false},"excerpt":{"rendered":"

Here you will find Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. Extra Questions for Class 10 Maths Introduction to Trigonometry with Answers Solutions Extra Questions for Class 10 Maths Chapter 8 …<\/p>\n

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