{"id":14286,"date":"2021-07-29T17:01:19","date_gmt":"2021-07-29T11:31:19","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=14286"},"modified":"2023-03-29T17:56:37","modified_gmt":"2023-03-29T12:26:37","slug":"ap-10th-class-maths-important-questions-chapter-11","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-10th-class-maths-important-questions-chapter-11\/","title":{"rendered":"AP 10th Class Maths Important Questions Chapter 11 Trigonometry"},"content":{"rendered":"

These AP 10th Class Maths Chapter Wise Important Questions<\/a> Chapter 11 Trigonometry will help students prepare well for the exams.<\/p>\n

AP State Syllabus 10th Class Maths 11th Lesson Important Questions and Answers Trigonometry<\/h2>\n

Question 1.
\nIf sin A = cos A, then find the value of A.
\nSolution:
\nIf sin A = cos A
\n\\(\\frac{\\sin A}{\\cos A}=\\frac{\\cos A}{\\cos A}\\) = 1
\n\u21d2 tan A = 1 = tan 45\u00b0 \u21d2 A = 45\u00b0
\n(OR)
\nsin 45\u00b0 = cos 45\u00b0
\n\\(\\frac{1}{\\sqrt{2}}=\\frac{1}{\\sqrt{2}}\\)
\nSo, A = 45\u00b0<\/p>\n

Question 2.
\nFind the value of x, if 2 sinx = \u221a3
\nSolution:
\n2 sinx = \u221a3
\nsin x = \\(\\frac{\\sqrt{3}}{2}\\)
\nsin x = sin 60\u00b0
\n\u2234 x = 60\u00b0<\/p>\n

\"AP<\/p>\n

Question 3.
\nEvaluate
\n(i) cos 76\u00b0 – sin 14\u00b0,
\n(ii) \\(\\frac{\\tan 73^{\\circ}}{\\cot 17^{\\circ}}\\)
\nSolution:
\nTo find the values of the following :
\ni) cos 76\u00b0 – sin 14\u00b0
\nWe can write cos 76 as cos (90 – 14)
\n\u2234 cos 76 = cos (90 – 14) = sin 14
\n(\u2235 cos (90 – \u03b8) = sin \u03b8)
\n\u2234 cos 76\u00b0 – sin 14\u00b0
\n= sin 14\u00b0 – sin 14\u00b0 = 0 ………… (1)<\/p>\n

(ii) \\(\\frac{\\tan 73^{\\circ}}{\\cot 17^{\\circ}}\\) = \\(\\frac{\\tan (90-17)}{\\cot 17}\\)
\n= \\(\\frac{\\cot 17^{\\circ}}{\\cot 17^{\\circ}}\\) = 1
\n(\u2235 tan (90 – \u03b8) = cot \u03b8<\/p>\n

Question 4.
\nFind the value of tan2<\/sup>45\u00b0 + cot2<\/sup> 30\u00b0.
\nSolution:
\ntan2<\/sup> 45\u00b0 + cot2<\/sup>30\u00b0
\n= (1)2<\/sup> + (\u221a3)2<\/sup>
\n= 1 + 3 = 4<\/p>\n

Question 5.
\nFind the value of sin2<\/sup> 30\u00b0 + cos2<\/sup> 60\u00b0.
\nsin 30\u00b0 = \\(\\frac{1}{2}\\),cos 60\u00b0 = \\(\\frac{1}{2}\\)
\nsin2<\/sup> 30\u00b0 + cos2<\/sup> 60\u00b0 = (\\(\\frac{1}{2}\\))2<\/sup> + (\\(\\frac{1}{2}\\))2<\/sup>
\n= \\(\\frac{1}{4}\\) + \\(\\frac{1}{4}\\) = \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\)
\n\u2234 sin2<\/sup> 30\u00b0 + cos2<\/sup> 60\u00b0 = \\(\\frac{1}{2}\\)<\/p>\n

Question 6.
\nIf sin (A + B) = 1 and cos B = \\(\\frac{1}{2}\\), then find \u2220A and \u2220B. (0\u00b0 < A + B \u2264 90\u00b0)
\nSolution:
\nsin (A + B) = 1
\nsin (A + B) = sin 90\u00b0
\nA + B = 90\u00b0
\ncos B = \\(\\frac{1}{2}\\)
\ncos B – cos 60\u00b0
\n\u2234 \u2220B = 60\u00b0
\n\u2234 \u2220A = 90\u00b0 – 60\u00b0 = 30\u00b0<\/p>\n

Question 7.
\nSimplify cot2<\/sup>\u03b8 – \\(\\frac{1}{\\sin ^{2} \\theta}\\)
\nSolution:
\n\"AP<\/p>\n

Question 8.
\nDoes sin\u03b8 \\(\\frac{5}{3}\\) exist for an acute angle \u03b8?
\nGive reason.
\nSolution:
\nGiven ‘\u03b8’ is acute => 0\u00b0 < \u03b8 < 90\u00b0
\nSo sin 0\u00b0 = 0 and sin 90\u00b0 = 1
\nSo for 0\u00b0 < \u03b8 < 90\u00b0,
\nsin \u03b8 value lies in between zero and one.
\nSo sin \u03b8 value cannot be greater than 1.
\nSo sin\u03b8 = \\(\\frac{5}{3}\\) does not exist.<\/p>\n

\"AP<\/p>\n

Question 9.
\nIf sin x = \\(\\frac{3}{4}\\), then what is the value of cosec x?
\nSolution:
\nIf sin x = \\(\\frac{3}{4}\\)
\n\"AP<\/p>\n

Question 10.
\nAmong sin 90\u00b0, cos 90\u00b0, tan 90\u00b0, cot 90\u00b0, sec 90\u00b0 and cosec 90\u00b0; which is\/are not defined ?
\nSolution:
\nsin 90\u00b0 =1
\ncos 90\u00b0 = 0
\ntan 90\u00b0 = not defined
\ncot 90\u00b0 = 0
\nsec 90\u00b0 = not defined
\ncosec 90\u00b0 = 1
\n\u2234 tan 90\u00b0, sec 90\u00b0 are not defined.<\/p>\n

Question 11.
\nExpress (sec2<\/sup> x – 1) (cot2<\/sup> x).
\nSolution:
\n(sec2<\/sup> x – 1) (cot2<\/sup> x)
\n= (tan2<\/sup> x) (cot2<\/sup> x) [\u2235 sec2<\/sup>A – tan2<\/sup>A = 1 sec2<\/sup> A – 1 = tan2<\/sup> A]
\n= \\(\\frac{\\sin ^{2} x}{\\cos ^{2} x} \\cdot \\frac{\\cos ^{2} x}{\\sin ^{2} x}\\) [\u2235 tan A = \\(\\frac{\\sin \\mathrm{A}}{\\cos \\mathrm{A}}\\) cot A = \\(\\frac{\\cos \\mathrm{A}}{\\sin \\mathrm{A}}\\)]
\n= 1<\/p>\n

Question 12.
\nFind \u2220B, if tan (A – B) = \\(\\frac{1}{\\sqrt{3}}\\) and sinA = \\(\\frac{\\sqrt{3}}{2}\\). Also find cos B. (A, B < 90\u00b0)
\nSolution:
\ntan (A – B) = \\(\\frac{1}{\\sqrt{3}}\\)
\n\u2234 tan (A – B) = tan 30\u00b0
\n\u2234 A – B = 30\u00b0
\nsin A = \\(\\frac{\\sqrt{3}}{2}\\)
\nsin A = sin 60\u00b0
\n\u2234 A = 60\u00b0
\nsubstituting A – B = 30\u00b0
\n60\u00b0 – B = 30\u00b0
\nB = 30\u00b0<\/p>\n

Question 13.
\nIf tan A = \\(\\frac{1}{\\sqrt{3}}\\) and tan B = \u221a3 , then find sin A. cos B + cos A . sin B.
\n(A, B < 90\u00b0).
\nSolution:
\nGiven, tan A = \\(\\frac{1}{\\sqrt{3}}\\) = and tan B = \u221a3
\nGiven A, B < 90\u00b0
\nWe know \\(\\frac{1}{\\sqrt{3}}\\) = tan 30\u00b0
\nso tan A = \\(\\frac{1}{\\sqrt{3}}\\) = tan 30\u00b0
\n\u21d2 A = 30\u00b0 …………….. (1)
\nand tan B = \u221a3 = tan 60\u00b0
\n\u21d2 B = 30\u00b0 …………….. (2)
\nthen
\nsin A cos B + cos A sin B
\n= sin 30\u00b0 cos 60\u00b0 + cos 30\u00b0 sin 60\u00b0
\nWhich can be written as
\nsin (A + B) = sin (30+60) = sin 90 = 1
\n\u2234 sin A cos B + cos A sin B = 1
\n(or)
\nsin 30 . cos 60 + cos 30 . sin 60
\n= \\(\\frac{1}{2} \\cdot \\frac{1}{2}+\\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{3}}{2}\\) = \\(\\frac{1}{4}+\\frac{3}{4}=\\frac{4}{4}\\) = 1<\/p>\n

Question 14.
\nProve that
\ntan2<\/sup>A – sin2<\/sup>A = tan2<\/sup>A . sin2<\/sup>A.
\nSolution:
\ntan2<\/sup>A – sin2<\/sup>A
\n= \\(\\frac{\\sin ^{2} A}{\\cos ^{2} A}\\) – sin2<\/sup>A
\nsin2<\/sup>A (\\(\\frac{1}{\\cos ^{2} A}\\) – 1 )
\n= sin2<\/sup>A (sec2<\/sup>A – 1)
\n= sin2<\/sup>A. tan2<\/sup>A<\/p>\n

\"AP<\/p>\n

Question 15.
\nIf cos A = \\(\\frac{7}{25}\\), then find sin A and cosec A. What do you observe?
\nSolution:
\nGiven that cos A = 7\/25
\nsin2<\/sup> A = 1 – cos2<\/sup> A
\n\"AP
\nI observed that sin A . cosec A = 1.<\/p>\n

Question 16.
\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= cot[90 – [90 – (A – 18\u00b0)]]
\ntan 2A = tan [90 – (A – 18\u00b0)]
\n2A = 90-(A – 18\u00b0) = 90 – A + 18\u00b0
\n\u21d2 3A = 108\u00b0
\n\u2234 A = 36\u00b0<\/p>\n

Question 17.
\nIf 4 tan \u03b8 = 3, then find the value of sec \u03b8 and cosec \u03b8.
\nSolution:
\n4 tan \u03b8 = 3 \u21d2 tan \u03b8 = \\(\\frac{3}{4}\\)
\n\"AP
\nAC = \\(\\sqrt{\\mathrm{BC}^{2}+\\mathrm{AB}^{2}}\\) = \\(\\sqrt{4^{2}+3^{2}}\\) = 5
\nsec \u03b8 = \\(\\frac{5}{4}\\); cosec \u03b8 = \\(\\frac{5}{3}\\)<\/p>\n

Question 18.
\nIf tan 2A = cot (A – 27), where 2A is an acute angle, find the value of A.
\nSolution:
\nGiven that 2A is an acute angle.
\nWe know if tan = \u03b1 = cot \u03b2 (\u03b1, \u03b2 acute) then \u03b1 + \u03b2 = 90\u00b0
\n(\u2235 tan (90 – \u03b8) = cot \u03b8)
\nSo from given
\ntan 2A = cot (A – 27)
\n\u21d2 2A + A – 27 = 90
\n\u21d2 2A + A = 90 + 27
\n\u21d2 3 A = 117 \u21d2 A = \\(\\frac{117}{3}\\) = 39
\n\u2234 A = 39\u00b0<\/p>\n

Question 19.
\nProve that
\n\\(\\sqrt{\\frac{1+\\sin A}{1-\\sin A}}\\) = sec A + tan A
\nSolution:
\n\"AP<\/p>\n

Question 20.
\nEvaluate (sin x – cos x)2<\/sup> + (sin x + cos x)2<\/sup>.
\nSolution:
\n(sin x – cos x)2<\/sup> + (sin x + cos x)2<\/sup>
\n= sin2<\/sup> x + cos2<\/sup> x – 2 sin x cos x + sin2<\/sup> x + cos2<\/sup> x + 2 sin x cos x
\n= 2(sin2<\/sup>x + cos2<\/sup>x) = 2(1) = 2<\/p>\n

\"AP<\/p>\n

Question 21.
\ncos A = \\(\\frac{12}{13}\\), then, find sin A and tan A
\nSolution:
\nGiven, cos A = \\(\\frac{12}{13}\\)
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n132 = 122<\/sup> + x2<\/sup>2
\n169 = 144 + x2<\/sup>
\nx2<\/sup> = 25 \u21d2 x = 5
\n\"AP
\n\"AP<\/p>\n

Question 22.
\nIf 4 sin2<\/sup> \u03b8 – 1 = 0 then find \u03b8’ (\u03b8 < 90) also, find the value of \u03b8 and the value of cos2<\/sup> \u03b8 + tan2<\/sup> \u03b8
\nSolution:
\nGiven 4 2<\/sup> \u03b8 – 1 = 0 \u21d2 4 2<\/sup> \u03b8 = 1
\n2<\/sup> \u03b8 = \\(\\frac{1}{4}\\) \u21d2 sin \u03b8 = \u00b1 \\(\\sqrt{\\frac{1}{4}=} \\pm \\frac{1}{2}\\)
\nGiven \u03b8 is less than 90\u00b0
\nsin \u03b8 = \\(\\frac{1}{2}\\)
\nsin \u03b8 = sin30\u00b0 \u21d2 \u03b8 = 30\u00b0
\ncos \u03b8 = cos = 30\u00b0 \\(\\frac{\\sqrt{3}}{2}\\)
\ntan \u03b8 = tan 30\u00b0 = \\(\\frac{1}{\\sqrt{3}}\\)
\ncos2<\/sup> \u03b8 + tan2<\/sup> \u03b8 = cos2<\/sup> 30\u00b0 + tan2<\/sup> 30\u00b0
\n= (\\(\\frac{\\sqrt{3}}{2}\\))2<\/sup> + (\\(\\frac{1}{\\sqrt{3}}\\))2<\/sup>
\n= \\(\\frac{3}{4}+\\frac{1}{3}\\) = \\(\\frac{9+4}{12}\\) = \\(\\frac{13}{12}\\)<\/p>\n

Question 23.
\nProve that: (sin \u03b8 – cosec \u03b8)2<\/sup> + (cos \u03b8 – sec \u03b8)2<\/sup> = cot2<\/sup> \u03b8 + tan2<\/sup> \u03b8 – 1
\nSolution:
\n(sin \u03b8 – cosec \u03b8)2<\/sup> + (cos \u03b8 – sec \u03b8)2<\/sup>
\n= sin2<\/sup>\u03b8 + cosec2<\/sup>\u03b8 – 2 sin\u03b8 . cosec\u03b8 . + cos2<\/sup>\u03b8 + sec2<\/sup>\u03b8 – 2 cos\u03b8 . sec \u03b8
\n= (sin2<\/sup>\u03b8 + cos2<\/sup>\u03b8) + cosec2<\/sup>\u03b8 + sec2<\/sup>\u03b8 – 2 – 2
\n= 1 + (1 + cot2<\/sup>\u03b8) + (1 + tan2<\/sup>\u03b8) – 2 – 2
\n= cot2<\/sup>\u03b8 + tan2<\/sup>\u03b8 + 3 – 4
\n= cot2<\/sup>\u03b8 + tan2<\/sup>\u03b8 – 1<\/p>\n

Question 24.
\nIf cosec \u03b8 + cot \u03b8 = p, show that \\(\\frac{p^{2}+1}{p^{2}-1}\\) = sec \u03b8.
\nSolution:
\ncosec \u03b8 – cot \u03b8 = p
\n\u2234 cosec \u03b8 – cot \u03b8 = \\(\\frac{1}{\\mathrm{p}}\\) ……………….(1)
\ncosec \u03b8 + cot \u03b8 = p …………..(2)
\n(1) + (2), We get
\n2 cosec \u03b8 = p + \\(\\frac{1}{\\mathrm{p}}\\) = \\(\\frac{p^{2}+1}{p}\\) …………..(3)
\ncosec \u03b8 – cot \u03b8 = p
\ncosec \u03b8 – cot \u03b8 = \\(\\frac{1}{\\mathrm{p}}\\)
\n(1) – (2), We get
\n\"AP
\nHence proved.<\/p>\n

Question 25.
\nProve that (1 + tan2<\/sup>\u03b8) + (1 + \\(\\frac{1}{\\tan ^{2} \\theta}\\)) = \\(\\frac{1}{\\sin ^{2} \\theta-\\sin ^{4} \\theta}\\)
\nSolution:
\n\"AP
\n\"AP<\/p>\n

Question 26.
\nIf sec \u03b8 + tan \u03b8 = p, then prove that sin \u03b8 = \\(\\frac{\\mathbf{p}^{2}-1}{\\mathbf{p}^{2}+1}\\)
\nSolution:
\nsec \u03b8 + tan \u03b8 = p
\nsec2 \u03b8 – tan2 \u03b8 = 1
\n(sec \u03b8 + tan \u03b8) (sec \u03b8 – tan \u03b8) = 1
\np. (sec \u03b8 – tan \u03b8) = 1
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 27.
\nIf cot \u03b8 = \\(\\frac{7}{8}\\) then,
\nEvaluate :
\ni) \\(\\frac{(1+\\sin \\theta)(1-\\sin \\theta)}{(1+\\cos \\theta)(1-\\cos \\theta)}\\)
\nii) \\(\\frac{1+\\cos \\theta}{\\sin \\theta}\\)
\nSolution:
\n\"AP<\/p>\n

Question 28.
\nShow that
\nsec2<\/sup> \u03b8 + cosec2<\/sup> \u03b8 = sec2<\/sup> \u03b8 . cosec2<\/sup> \u03b8.
\nSolution:
\nsec2<\/sup> \u03b8 + cosec2<\/sup> \u03b8
\n\"AP<\/p>\n","protected":false},"excerpt":{"rendered":"

These AP 10th Class Maths Chapter Wise Important Questions Chapter 11 Trigonometry will help students prepare well for the exams. AP State Syllabus 10th Class Maths 11th Lesson Important Questions and Answers Trigonometry Question 1. If sin A = cos A, then find the value of A. Solution: If sin A = cos A = … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":35710,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[12],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/07\/Chapter-11-Trigonometry.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/14286"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=14286"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/14286\/revisions"}],"predecessor-version":[{"id":35711,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/14286\/revisions\/35711"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/35710"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=14286"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=14286"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=14286"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}