2<\/sup> (x + \\(\\frac{2 \\pi}{3}\\)) = \\(\\frac{3}{2}\\)
\nSolution:<\/p>\n
<\/p>\n
<\/p>\n
Question 21.
\nIf x, y and z are in A.P., prove that y = \\(\\frac{\\sin x-\\sin z}{\\cos z-\\cos x}\\).
\nSolution:
\nSince x, y, z are in A.P.
\n\u2234 y – x = z – y
\n\u21d2 2y = x + z<\/p>\n
<\/p>\n
Question 22.
\nIf sin 2x = \u03bb sin 2y, prove that \\(\\frac{\\tan (x+y)}{\\tan (x-y)}=\\frac{\\lambda+1}{\\lambda-1}\\).
\nSolution:
\nGiven sin 2x = \u03bb sin 2y<\/p>\n
<\/p>\n
Question 23.
\nProve that \\(\\frac{1-\\sin 2 x}{1+\\sin 2 x}=\\tan ^2\\left(\\frac{\\pi}{4}-x\\right)\\).
\nSolution:<\/p>\n
<\/p>\n
<\/p>\n
Question 24.
\nProve that \\(\\sin ^4 \\frac{\\pi}{8}+\\sin ^4 \\frac{3 \\pi}{8}+\\sin ^4 \\frac{5 \\pi}{8}+\\sin ^4 \\frac{7 \\pi}{8}=\\frac{3}{2}\\).
\nSolution:<\/p>\n
<\/p>\n
Question 25.
\nGiven that sin x = – \\(\\frac{3}{5}\\), cos y = – \\(\\frac{5}{13}\\) and x is in the same quadrant asy, evaluate without using tables :
\n(i) cos (x – y)
\n(ii) tan (x + y)
\n(iii) cos \\(\\frac{x}{2}\\)
\nSolution:
\nGiven sin x = – \\(\\frac{3}{5}\\)
\nand cos y = – \\(\\frac{5}{13}\\)
\nSince sin x and cos y both are negative.
\n\u2234x and y lies in 3rd quadrant as x and y lies in same quadrant.
\nSo \\(\\frac{x}{2}\\), \\(\\frac{y}{2}\\) lies in second qu1drant.
\n\u2234 cos x = – \\(\\sqrt{1-\\sin ^2 x}\\)
\n= \\(-\\sqrt{1-\\left(-\\frac{3}{5}\\right)^2}\\)
\n= \\(-\\sqrt{1-\\frac{9}{25}}\\)
\n= \\(-\\sqrt{\\frac{16}{25}}=-\\frac{4}{5}\\)
\nand sin y = – \\(\\sqrt{1-\\cos ^2 y}\\)
\n= \\(-\\sqrt{1-\\left(\\frac{5}{13}\\right)^2}\\)
\n= \\(-\\sqrt{1-\\frac{25}{169}}\\)
\n= \\(-\\sqrt{\\frac{144}{169}}=-\\frac{12}{13}\\)<\/p>\n
(i) cos (x – y) = cos x cos y + sin x sin y
\n= \\(\\left(-\\frac{4}{5}\\right)\\left(-\\frac{5}{13}\\right)+\\left(-\\frac{3}{5}\\right)\\left(-\\frac{12}{13}\\right)\\)
\n= \\(\\frac{20+36}{65}=\\frac{56}{65}\\)<\/p>\n
(ii) tan (x + y) = \\(\\frac{\\tan x+\\tan y}{1-\\tan x \\tan y}\\)
\n= \\(\\frac{\\frac{3}{4}+\\frac{12}{5}}{1-\\frac{3}{4} \\times \\frac{12}{5}}\\)
\n= \\(\\frac{\\frac{15+48}{20}}{1-\\frac{36}{20}}\\)
\n= \\(\\frac{63}{-16}\\)
\n[tan x = \\(=\\frac{\\sin x}{\\cos x}\\)
\n= \\(\\frac{-\\frac{3}{5}}{-\\frac{4}{5}}=\\frac{3}{4}\\)
\nand tan y = \\(\\frac{\\sin y}{\\cos y}\\)
\n= \\(\\frac{-\\frac{12}{13}}{-\\frac{5}{13}}=\\frac{12}{5}\\)]<\/p>\n
(iii) since \\(\\frac{x}{2}\\) lies in second quadrant
\n\u2234 cos x < 0
\nThus, cos \\(\\frac{x}{2}\\) = – \\(\\sqrt{\\frac{1+\\cos x}{2}}\\)
\n= – \\(\\sqrt{\\frac{1+\\left(-\\frac{4}{5}\\right)}{2}}\\)
\n= \\(-\\sqrt{\\frac{1}{10}}=-\\frac{1}{\\sqrt{10}}\\)<\/p>\n
<\/p>\n
Question 26.
\nProve that :
\n(i) \\(\\frac{1-\\cos x+\\cos y-\\cos (x+y)}{1+\\cos x-\\cos y-\\cos (x+y)}=\\tan \\frac{x}{2} \\cot \\frac{y}{2}\\)
\n(ii) \\(\\frac{\\cos ^3 x-\\cos 3 x}{\\cos x}+\\frac{\\sin ^3 x+\\sin 3 x}{\\sin x}\\) = 3.
\nSolution:
\n(i)<\/p>\n
<\/p>\n
(ii)<\/p>\n
<\/p>\n
Question 27.
\nProve that \\(\\frac{1}{\\sin 10^{\\circ}}-\\frac{\\sqrt{3}}{\\cos 10^{\\circ}}\\) = 4.
\nSolution:<\/p>\n
<\/p>\n
<\/p>\n
Question 28.
\nShow that never lies between \\(\\frac{1}{3}\\) and 3.
\nSolution:
\nLet y = \\(\\frac{\\tan 3 x}{\\tan x}\\)
\n= \\(\\frac{3 \\tan x-\\tan ^3 x}{\\tan x\\left(1-3 \\tan ^2 x\\right)}\\)
\n= \\(\\frac{3-\\tan ^2 x}{1-3 \\tan ^2 x}\\)
\n\u21d2 y (1 – 3 tan2<\/sup> x) = 3 – tan2<\/sup> x
\n\u21d2 tan2<\/sup> x (1 – 3y) = 3 – y
\n\u21d2 tan2<\/sup> x = \\(\\frac{3-y}{1-3 y}\\)
\nBut tan2<\/sup> x \u2265 0 \u2200 x \u2208 R
\n\u21d2 \\(\\frac{3-y}{1-3 y}\\) \u2265 0
\n\u21d2 (3 – y) (1 – 3y) \u2265 0
\n[\u2235 (1 – 3y)2<\/sup> \u2265 0 \u2200 x \u2208 R]
\n\u21d2 (y – 3) (3y – 1) \u2265 0
\n\u21d2 (y – 3) (y – \\(\\frac{1}{3}\\)) \u2265 0
\n\u21d2 y \u2265 3 or y \u2264 \\(\\frac{1}{3}\\)
\n[\u2235 (x – a) (x – b) \u2265 0
\nand a > b = x \u2265 a or x \u2264 b]
\nThus y i.e. \\(\\frac{\\tan 3 x}{\\tan x}\\) never lies between \\(\\frac{1}{3}\\) and 3.<\/p>\nQuestion 29.
\nSolve the following equations :
\n(i) tan 2x = – cot (x + \\(\\frac{\\pi}{6}\\))
\n(ii) cot2<\/sup> x + 3 cosec x + 3 = 0
\n(iii) 4 sin2<\/sup> x + \u221a3 = 2 (1 + \u221a3) sin x
\n(iv) tan2<\/sup> x – (1 + \u221a3) tan x + \u221a3 = 0
\n(v) cos 2x – cos 8x + cos 6x = 1
\n(vi) tan (\\(\\frac{\\pi}{4}\\) + x) + tan (\\(\\frac{\\pi}{4}\\) – x) = 4
\n(vii) cosec x = 1 + cot x
\nSolution:
\n(i) tan 2x = – cot (x + \\(\\frac{\\pi}{6}\\))
\n\u21d2 tan 2x = tan \\(\\left(\\frac{\\pi}{2}+x+\\frac{\\pi}{6}\\right)\\)
\n= tan \\(\\left(\\frac{2 \\pi}{3}+x\\right)\\)
\n\u21d2 2x = n\u03c0 + \\(\\frac{2 \\pi}{3}\\) + x ; n \u2208 I
\n[\u2235 tan \u03b8 = tan \u03b1
\n\u21d2 \u03b8 = n\u03c0 + \u03b1, n \u2208 I]
\n\u21d2 x = n\u03c0 + \\(\\frac{2 \\pi}{3}\\) ; n \u2208 I<\/p>\n(ii) Given eqn. be,
\ncot2<\/sup> x + 3 cosec x + 3 = 0
\n\u21d2 cosec2<\/sup> x – 1 + 3 cosec x + 3 = 0
\n\u21d2 cosec2<\/sup> x + 3 cosec x + 2 = 0
\n\u21d2 cosec x = \\(\\frac{-3 \\pm \\sqrt{9-8}}{2}\\)
\n= \\(\\frac{-3 \\pm 1}{2}\\)
\n= – 2, – 1
\neither cosec x = – 2 or cosec x = – 1
\nsin x = – \\(\\frac{1}{2}\\)
\nor sin x = – 1
\n\u21d2 sin x = – sin \\(\\frac{\\pi}{6}\\)
\nor sin x = sin (- \\(\\frac{\\pi}{2}\\))
\n\u21d2 x = n\u03c0 + (- 1)n<\/sup> (- \\(\\frac{\\pi}{6}\\))
\nor x = n\u03c0 + (- 1)n<\/sup> (- \\(\\frac{\\pi}{2}\\))
\nHence required solutions of given eqn. are
\nx = n\u03c0 – (- 1)n<\/sup> (- \\(\\frac{\\pi}{6}\\)) or n\u03c0 – (- 1)n<\/sup> (- \\(\\frac{\\pi}{2}\\))<\/p>\n(iii) Given eqn. be,
\n4 sin2<\/sup> x + \u221a3 = 2 (1 + \u221a3) sin x<\/p>\n
<\/p>\n
(iv) Given eqn. be,
\ntan2<\/sup> x – (1 + \u221a3) tan x + \u221a3 = 0
\nwhich is quadratic in tan x.
\n\u2234 tan x = \\(\\frac{(1+\\sqrt{3}) \\pm \\sqrt{(1+\\sqrt{3})^2-4 \\sqrt{3}}}{2}\\)
\n= \\(\\frac{(1+\\sqrt{3}) \\pm \\sqrt{(1-\\sqrt{3})^2}}{2}\\)
\n= \\(\\frac{(1+\\sqrt{3}) \\pm(1-\\sqrt{3})}{2}\\)
\n= 1, \u221a3
\neither tan x = 1 or tan x = \u221a3
\ntan x = 1 = tan \\(\\frac{\\pi}{4}\\)
\ntan x = \u221a3 = tan \\(\\frac{\\pi}{3}\\)
\n\u21d2 x = n\u03c0 + \\(\\frac{\\pi}{4}\\)
\nor x = n\u03c0 + \\(\\frac{\\pi}{3}\\) \u2200 n \u2208 I
\nHence required solutions are n\u03c0 + \\(\\frac{\\pi}{4}\\), n\u03c0 + \\(\\frac{\\pi}{3}\\) ; n \u2208 I<\/p>\n(v) Given eqn.be
\ncos 2x – cos 8x + cos 6x = 1
\n\u21d2cos 2x – cos 8x – (1 – cos 6x) = 0
\n\u21d2 2 sin \\(\\left(\\frac{2 x+8 x}{2}\\right)\\) sin \\(\\left(\\frac{8 x-2 x}{2}\\right)\\) – 2 sin2<\/sup> 3x = 0
\n\u21d2 2 sin 5x sin 3x – 2 sin2<\/sup> 3x = 0
\n\u21d2 2 sin 3x [sin 5x – sin 3x] = 0
\n\u21d2 2 sin 3x (2 cos 4x sin x) = 0
\neither sin 3x = 0 or cos 4x =0 or sin x = 0
\n3x = n\u03c0 or 4x = (2n + 1) \\(\\frac{\\pi}{2}\\)
\nor x = n\u03c0 ; n \u2208 I
\n\u21d2 x = \\(\\frac{n \\pi}{3}\\) or x = (2n + 1) \\(\\frac{\\pi}{8}\\)
\nor x = n\u03c0 ; n \u2208 I
\nSince the solution x = n\u03c0
\ni.e. x = \u03c0, 2\u03c0, 3\u03c0 ……………….. are included in the
\nsolution x = \\(\\frac{n \\pi}{3}\\) for n = 3, 6, 9 ……………..
\nHence, the required solutions are
\nx = n\u03c0, (2n + 1) \\(\\frac{\\pi}{8}\\) \u2200 n \u2208 I<\/p>\n(vi) Given tan \\(\\left(\\frac{\\pi}{4}+x\\right)\\) + tan \\(\\left(\\frac{\\pi}{4}-x\\right)\\) = 4
\n\u21d2 \\(\\frac{\\tan \\frac{\\pi}{4}+\\tan x}{1-\\tan \\frac{\\pi}{4} \\tan x}+\\frac{\\tan \\frac{\\pi}{4}-\\tan x}{1+\\tan \\frac{\\pi}{4} \\tan x}\\) = 4
\n\u21d2 \\(\\frac{1+\\tan x}{1-\\tan x}+\\frac{1-\\tan x}{1+\\tan x}\\) = 4
\n\u21d2 (1 + tan x)2<\/sup> + (1 – tan x)2<\/sup> = 4 (1 – tan2<\/sup> x)
\n\u21d2 2 tan2<\/sup> x + 2 = 4 (1 – tan2<\/sup> x)
\n\u21d2 6 tan2<\/sup> x = 2
\n\u21d2 tan2<\/sup> x = \\(\\frac{1}{3}=\\left(\\frac{1}{\\sqrt{3}}\\right)^2\\)
\n= tan2<\/sup> \\(\\frac{\\pi}{6}\\)
\n\u21d2 x = n\u03c0 \u00b1 \\(\\frac{\\pi}{6}\\) \u2200 n \u2208 I
\n[\u2235 tan2<\/sup> \u03b8 = tan2<\/sup> \u03b1
\n\u21d2 n\u03c0 \u00b1 \u03b1 \u2200 n \u2208 I]<\/p>\n(vii) Given cosec x = 1 + cot x
\n\u21d2 \\(\\frac{1}{\\sin x}=1+\\frac{\\cos x}{\\sin x}\\)
\n\u21d2 1 = sin x + cos x ; sin x \u2260 0 …………….(1)
\nwhich is of the form a cos x + b sin x = c
\ndividing throughout eqn. (1) by \\(\\sqrt{1^2+1^2}\\) i.e. \u221a2 ; we get
\n\\(\\frac{1}{\\sqrt{2}}\\) cos x + \\(\\frac{1}{\\sqrt{2}}\\) sin x = \\(\\frac{1}{\\sqrt{2}}\\)<\/p>\n
<\/p>\n
\u21d2 x = 2n\u03c0 + \\(\\frac{\\pi}{2}\\), 2n\u03c0 ; n \u2208 I
\nAs sin (2n\u2208) = 0 \u2200 n \u2208 I but sin x \u2260 0
\nHence the required solutions are 2n\u03c0 + \\(\\frac{\\pi}{2}\\) ; n \u2208 I<\/p>\n
Question 30.
\nIn any triangle ABC, prove that
\n(i) 2 (b cos2<\/sup> \\(\\frac{C}{2}\\) + c cos2<\/sup> \\(\\frac{B}{2}\\)) = a + b + c
\n(ii) \\(\\frac{\\sin \\mathrm{A}}{\\sin (A+B)}=\\frac{a}{c}\\)
\n(iii) \\(\\frac{a-b}{a+b}=\\frac{\\tan \\frac{A-B}{2}}{\\tan \\frac{A+B}{2}}\\)
\n(iv) \\(\\frac{b+c}{b-c}=\\cot \\frac{A}{2} \\cot \\frac{B-C}{2}\\)
\n(v) \\(\\frac{1+\\cos (\\mathrm{A}-\\mathrm{B}) \\cos \\mathrm{C}}{1+\\cos (\\mathrm{A}-\\mathrm{C}) \\cos \\mathrm{B}}=\\frac{a^2+b^2}{a^2+c^2}\\)
\nSolution:
\n(i) L.H.S. = 2 \\(\\left(b \\cos ^2 \\frac{\\mathrm{C}}{2}+c \\cos ^2 \\frac{\\mathrm{B}}{2}\\right)\\)
\n= 2 \\(\\left[b\\left(\\frac{1+\\cos \\mathrm{C}}{2}\\right)+c\\left(\\frac{1+\\cos \\mathrm{B}}{2}\\right)\\right]\\)
\n= [b + c + b cos C + c cos B]
\n= b + c + a [using projection formulae]
\n= R.H.S.<\/p>\n(ii) L.H.S. = \\(\\frac{\\sin A}{\\sin (A+B)}\\)
\n= \\(\\frac{\\sin A}{\\sin (\\pi-C)}\\)
\n[\u2235 A + B + C = \u03c0
\n\u21d2 A + B = \u03c0 – C]
\n= \\(\\frac{\\sin \\mathrm{A}}{\\sin \\mathrm{C}}=\\frac{a}{c}\\)
\n[by sine formula, \\(\\frac{a}{\\sin \\mathrm{A}}=\\frac{b}{\\sin \\mathrm{B}}=\\frac{c}{\\sin \\mathrm{C}}\\)
\n\u21d2 \\(\\frac{\\sin \\mathrm{A}}{\\sin \\mathrm{C}}=\\frac{a}{c}\\)]
\n= R.H.S.<\/p>\n
(iii)<\/p>\n
<\/p>\n
(iv)<\/p>\n
<\/p>\n
(v)<\/p>\n
<\/p>\n
Question 31.
\nIn a triangle ABC, if cos A + cos C = sin B, then prove that it is right anglcd triangle.
\nSolution:
\nGiven cos A + cos C = sin B<\/p>\n
<\/p>\n
either sin \\(\\frac{B}{2}\\) = 0 or cos C = 0 or cos A = 0
\ni.e. \\(\\frac{B}{2}\\) = 0 or C = 90\u00b0 or A = 90\u00b0
\ni.e. B = 0 or C = 90\u00b0 or A = 90\u00b0
\neither C = 90\u00b0 or A = 90\u00b0 [\u2235 B \u2260 0]
\nHence \u2206 is right angled triangle.<\/p>\n","protected":false},"excerpt":{"rendered":"
Effective ML Aggarwal Class 11 Solutions ISC Chapter 3 Trigonometry Chapter Test can help bridge the gap between theory and application. ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test Question 1. Find the radian measure of an angle (internal) of a regular (i) pentagon (ii) hexagon (iii) polygon of n …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170736"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=170736"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170736\/revisions"}],"predecessor-version":[{"id":170756,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170736\/revisions\/170756"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170736"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170736"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170736"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}