{"id":170736,"date":"2024-06-06T14:51:26","date_gmt":"2024-06-06T09:21:26","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170736"},"modified":"2024-06-06T14:51:26","modified_gmt":"2024-06-06T09:21:26","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-3-chapter-test","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-3-chapter-test\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test"},"content":{"rendered":"

Effective ML Aggarwal Class 11 Solutions ISC<\/a> Chapter 3 Trigonometry Chapter Test can help bridge the gap between theory and application.<\/p>\n

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test<\/h2>\n

Question 1.
\nFind the radian measure of an angle (internal) of a regular
\n(i) pentagon
\n(ii) hexagon
\n(iii) polygon of n sides.
\nSolution:
\nWe know that interior angle of polygon with n sides = \\(\\left(\\frac{2 n-4}{n}\\right)\\) \u00d7 90\u00b0
\nor \\(\\left(\\frac{2 n-4}{n}\\right) \\frac{\\pi}{2}\\) rad
\n(i) In pentagon, n = 5
\n\u2234 interior angle = \\(\\left(\\frac{2 \\times 5-4}{5}\\right) \\frac{\\pi}{2}\\) rad
\n= \\(\\frac{6}{5} \\times \\frac{\\pi}{2}\\) radian
\n= \\(\\frac{3 \\pi}{5}\\) rad.<\/p>\n

(ii) In case of hexagon, n = 6
\n\u2234 required interior angle = \\(\\left(\\frac{2 \\times 6-4}{6}\\right) \\frac{\\pi}{2}\\) rad
\n= \\(\\left(\\frac{4}{3} \\times \\frac{\\pi}{2}\\right)\\) rad
\n= \\(\\frac{2 \\pi}{3}\\) rad.<\/p>\n

(iii) In case of polygon of n sides
\n\u2234 required interior angle = \\(\\left(\\frac{2 n-4}{n}\\right) \\frac{\\pi}{2}\\)
\n= \\(\\left(\\frac{n-2}{n}\\right)\\) \u03c0 rad<\/p>\n

\"ML<\/p>\n

Question 2.
\nFind the value of sin (- \\(\\frac{41 \\pi}{4}\\)).
\nSolution:
\nsin (- \\(\\frac{4 \\pi}{4}\\)) = – sin (\\(\\frac{41 \\pi}{4}\\))
\n[\u2235 sin (- \u03b8) = – sin \u03b8]
\n= – sin \\(\\left(\\frac{40 \\pi+\\pi}{4}\\right)\\)
\n= – sin \\(\\left(10 \\pi+\\frac{\\pi}{4}\\right)\\)
\n= – sin \\(\\frac{\\pi}{4}\\)
\n= – \\(\\)
\n[.. sin(2nlt+0)sinOVnE 1]<\/p>\n

Question 3.
\nProve that :
\n(i) sin2<\/sup> \\(\\frac{\\pi}{6}\\), sin2<\/sup> \\(\\frac{\\pi}{4}\\), sin2<\/sup> \\(\\frac{\\pi}{3}\\) are in A.P.
\n(ii) tan2<\/sup> \\(\\frac{\\pi}{6}\\) , tan2<\/sup> \\(\\frac{\\pi}{4}\\), tan2<\/sup> \\(\\frac{\\pi}{3}\\) are in GP.
\nSolution:
\n(i)<\/p>\n

\"ML<\/p>\n

(ii)<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 4.
\nFind the value of m2<\/sup> sin \\(\\frac{1}{2}\\) \u03c0 – n2<\/sup> sin \\(\\frac{3}{2}\\) \u03c0 + 2mn sec \u03c0.
\nSolution:
\nm2<\/sup> sin \\(\\frac{1}{2}\\) \u03c0 – n2<\/sup> sin \\(\\frac{3}{2}\\) \u03c0 + 2mn sec \u03c0
\n= m2<\/sup> \u00d7 1 – n2<\/sup> sin (\u03c0 + \\(\\frac{\\pi}{2}\\)) + 2mn \\(\\frac{1}{\\cos \\pi}\\)
\n= m2<\/sup> – n2<\/sup> (9- 1) + 2mn (- 1)
\n= m2<\/sup> + n2<\/sup> – 2mn
\n= (m – n)2<\/sup><\/p>\n

Question 5.
\nWhat is the maximum value of 3 – 7 cos 5x?
\nSolution:
\nSince – 1 \u2264 cos 5 x \u2264 1
\n\u21d2 + 7 \u2265 – 7 cos 5x \u2265 – 7
\n\u21d2 – 7 \u2264 – 7 cos 5x \u2264 7
\n\u21d2 3 – 7 \u2264 3 – 7 cos 5x \u2264 3 + 7
\n\u21d2 – 4 \u2264 3 – 7 cos 5x \u2264 10
\nThus, max. value of 3 – 7 cos 5x = 10<\/p>\n

Question 6.
\nWhat is the minimum value of 4 + 5 sin (3x – 2) ?
\nSolution:
\nSince – 1 \u2264 sin (3x – 2) \u2264 1
\n[\u2235 |sin \u03b8| \u2264 1 \u2200 \u03b8]
\n\u21d2 – 5 \u2264 5 sin (3x – 2) \u2264 5
\n\u21d2 4 – 5 \u2264 4 + 5 sin (3x – 2) \u2264 4 + 5
\n\u21d2 – 1 \u2264 4 + 5 sin (3x – 2) \u2264 9
\nThus, min. value of 4 + 5 sin (3x – 2) = – 1.<\/p>\n

\"ML<\/p>\n

Question 7.
\nWhat is the maximum value of sin x cos x?
\nSolution:
\nLet f(x) = sin x cos x
\n= \\(\\frac{1}{2}\\) (2 sin x cos x)
\n= \\(\\frac{\\sin 2 x}{2}\\)
\nsince – 1 \u2264 sin 2x \u2264 1
\n\u21d2 – \\(\\frac{1}{2}\\) \u2264 \\(\\frac{1}{2}\\) sin 2x \u2264 \\(\\frac{1}{2}\\)
\n\u21d2 – \\(\\frac{1}{2}\\) \u2264 f(x) \u2264 \\(\\frac{1}{2}\\)
\nThus, max. value of f (x) = \\(\\frac{1}{2}\\).<\/p>\n

Question 8.
\nWhat is the maximum value of 3 sin x – 4 sin3<\/sup> x?
\nSolution:
\nLet f (x) = 3 sin x – 4 sin3<\/sup> x
\n= sin 3x
\nsince – 1 \u2264 sin 3x \u2264 1
\nThus, max. value of f (x) = 1.<\/p>\n

Question 9.
\nWhat is the minimum value of 3 cos x – 4 cos3<\/sup> x?
\nSolution:
\nLet f (x) = 3 cos x – 4 cos3<\/sup> x
\n= – cos 3x
\nsince, – 1 \u2264 cos 3x \u2264 1
\n\u21d2 1 \u2265 – cos3x \u2265 – 1
\n\u21d2 – 1 \u2264 \u2014 cos 3x \u2264 1
\n\u21d2 – 1 \u2264 f (x) \u2264 1
\nThus, min value of f (x) = – 1.<\/p>\n

Question 10.
\nWhat is the least value of 2 sin2<\/sup> x + 3 cos2<\/sup> x?
\nSolution:
\nLet f (x) = 2 sin2<\/sup> x + 3 cos2<\/sup> x
\n= 2 (sin2<\/sup> x + cos2<\/sup>x) + cos2<\/sup> x
\n= 2 + cos2<\/sup> x
\nsince cos2<\/sup> x \u2265 0
\n\u21d2 2 + cos2<\/sup> x \u2265 2
\n\u21d2 f (x) \u2265 2
\nThus, least value of f (x) = 2.<\/p>\n

\"ML<\/p>\n

Question 11.
\nWhat is the maximum value of sin x + cos x?
\nSolution:
\nLet f (x) = sin x + cos x<\/p>\n

\"ML<\/p>\n

Question 12.
\nWhat is the minimum value of sin x – cos x?
\nSolution:
\nLet f (x) = sin x – cos x
\n= \u221a2 (\\(\\frac{1}{\\sqrt{2}}\\) sin x – \\(\\frac{1}{\\sqrt{2}}\\) cos x)
\n= \u221a2 [cos \\(\\frac{\\pi}{4}\\) sin x – sin \\(\\frac{\\pi}{4}\\) cos x]
\n= \u221a2 sin (x – \\(\\frac{\\pi}{4}\\))
\nsince – 1 \u2264 sin (x- \\(\\frac{\\pi}{4}\\)) \u2264 1
\n\u21d2 – \u221a2 \u2264 \u221a2 sin (x – \\(\\frac{\\pi}{4}\\)) \u2264 \u221a2
\n\u21d2 – \u221a2 \u2264 f (x) \u2264 \u221a2
\nThus, min value of f (x) = – \u221a2<\/p>\n

Question 13.
\nIf sin 2x = cos 3x and 0 \u2264 x < \\(\\frac{\\pi}{2}\\), then find the value of x.
\nSolution:
\nGiven sin 2x = cos 3x<\/p>\n

\"ML<\/p>\n

Question 14.
\nA railway carriage is travelling along a circular railway track of radius 1500 metres with a speed of 66 km \/ hour. Find the angle in degrees turned by the engine in 10 seconds.
\nSolution:
\nGiven radius of circular railway track = r
\n= 1500 cm
\ndistance covered by a railway carriage in 1 hour = 66 km
\nSo distance covered by a railway carriage in 1 second = \\(\\frac{66 \\times 1000}{60 \\times 60}\\)
\nThus distance covered by a railway carriage in 10 seconds
\ns = \\(\\frac{10 \\times 60 \\times 1000}{60 \\times 60}\\) m
\n= \\(\\frac{550}{3}\\) m
\n\u2234 required angle turned by engine in 10 seconds \u03b8 = \\(\\frac{s}{r}\\)
\n\u2234 \u03b8 = \\(\\frac{\\frac{550}{3}}{1500}\\)
\n= \\(\\frac{550}{3 \\times 1500}=\\frac{11}{90}\\) rad
\n= \\(\\frac{11}{90} \\times \\frac{180^{\\circ}}{22}\\) \u00d7 7
\n[\u2235 \u03c0 rad = 180\u00b0]
\n= 7\u00b0<\/p>\n

\"ML<\/p>\n

Question 15.
\nIf tan x = \\(\\frac{a}{b}\\), show that \\(\\frac{a \\sin x-b \\cos x}{a \\sin x+b \\cos x}=\\frac{a^2-b^2}{a^2+b^2}\\).
\nSolution:
\nGiven tan x = \\(\\frac{a}{b}\\)
\nL.H.S. = \\(\\frac{a \\sin x-b \\cos x}{a \\sin x+b \\cos x}\\)
\n= \\(\\frac{a \\tan x-b}{a \\tan x+b}\\)
\n= \\(\\frac{a\\left(\\frac{a}{b}\\right)-b}{a\\left(\\frac{a}{b}\\right)+b}\\)
\n= \\(\\frac{a^2-b^2}{a^2+b^2}\\)<\/p>\n

Question 16.
\nIs the equation sec2<\/sup> x – sec x + 1 = 0 possible?
\nSolution:
\nGiven eqn. be.
\n6 sec2<\/sup> x – 5 sec x + 1 = 0
\n\u21d2 sec x = \\(\\)
\n= \\(\\)
\nwhich is not possible since |sec x| \u2265 1 \u2200 x
\nHence given eqn. (1) is not possible.<\/p>\n

Question 17.
\nShow that \u221a3 (tan 17\u00b0 – tan 140\u00b0) = 1 + tan 170\u00b0 tan 140\u00b0.
\nSolution:
\nNow \\(\\frac{\\tan 170^{\\circ}-\\tan 140^{\\circ}}{1+\\tan 170^{\\circ} \\tan 140^{\\circ}}\\) = tan (170\u00b0 – 140\u00b0)
\n[\u2235 tan (A – B) = \\(\\frac{\\tan A-\\tan B}{1+\\tan A \\tan B}\\)]
\n= tan 30\u00b0 = \\(\\frac{1}{\\sqrt{3}}\\)
\n\u21d2 \u221a3 (tan 17\u00b0 – tan 140\u00b0) = 1 + tan 170\u00b0 tan 140\u00b0<\/p>\n

\"ML<\/p>\n

Question 18.
\nIf A + B = 45\u00b0, prove that (1 + tan A) (1 + tan B) = 2. Hence, find the value of tan 22 \\(\\frac{1}{2}^{\\circ}\\).
\nSolution:
\nGiven A + B = 45\u00b0 ………………(1)
\nL.H.S. = (1 + tan A) (1 + tan B)
\n= (1 + tan A) (1 + tan (45\u00b0 – A)) [using (1)]
\n= (1 + tan A) \\(\\left[1+\\frac{\\tan 45^{\\circ}-\\tan \\mathrm{A}}{1+\\tan 45^{\\circ} \\tan \\mathrm{A}}\\right]\\)
\n= (1 + tan A) \\(\\left[1+\\frac{1-\\tan A}{1+\\tan A}\\right]\\)
\n= (1 + tan A) \\(\\left[\\frac{1+\\tan A+1-\\tan A}{1+\\tan A}\\right]\\)
\n= 2 = R.H.S.<\/p>\n

Question 19.
\nIf tan y = \\(\\frac{Q \\sin x}{P+Q \\cos x}\\), prove that tan (x – y) = \\(\\frac{P \\sin x}{Q+P \\cos x}\\).
\nSolution:
\nGiven tan y = \\(\\frac{Q \\sin x}{P+Q \\cos x}\\) ………………..(1)<\/p>\n

\"ML<\/p>\n

Question 20.
\nProve that cos2<\/sup> (x – \\(\\frac{2 \\pi}{3}\\)) + cos2<\/sup> (x + \\(\\frac{2 \\pi}{3}\\)) = \\(\\frac{3}{2}\\)
\nSolution:<\/p>\n

\"ML<\/p>\n

\"ML<\/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

\"ML<\/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

\"ML<\/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

\"ML<\/p>\n

\"ML<\/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

\"ML<\/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

\"ML<\/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

\"ML<\/p>\n

(ii)<\/p>\n

\"ML<\/p>\n

Question 27.
\nProve that \\(\\frac{1}{\\sin 10^{\\circ}}-\\frac{\\sqrt{3}}{\\cos 10^{\\circ}}\\) = 4.
\nSolution:<\/p>\n

\"ML<\/p>\n

\"ML<\/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>\n

Question 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

\"ML<\/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

\"ML<\/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

\"ML<\/p>\n

(iv)<\/p>\n

\"ML<\/p>\n

(v)<\/p>\n

\"ML<\/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

\"ML<\/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}]}}