{"id":17451,"date":"2022-05-25T19:30:12","date_gmt":"2022-05-25T14:00:12","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17451"},"modified":"2022-05-17T11:06:11","modified_gmt":"2022-05-17T05:36:11","slug":"mcq-questions-for-class-11-maths-chapter-3","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-maths-chapter-3\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Trigonometric Functions Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 3 Trigonometric Functions Objective Questions.<\/p>\n

Trigonometric Functions Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Trigonometric Functions Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Trigonometric Functions Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Trigonometric Functions Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nThe value of sin 15 + cos 15 is
\n(a) 1
\n(b) 1\/2
\n(c) \u221a3\/2
\n(d) \u221a3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) \u221a3\/2
\nGiven, sin 15 + cos 15
\n= sin 15 + cos(90 – 15)
\n= sin 15 + sin 15
\n= 2 \u00d7 sin 45 \u00d7 cos 30
\n= 2 \u00d7 (1\/\u221a2) \u00d7 (\u221a3\/2)
\n= \u221a3\/2<\/p>\n<\/details>\n

\n

Question 2.
\nThe value of tan A\/2 – cot A\/2 + 2cot A is
\n(a) 0
\n(b) 1
\n(c) -1
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 0
\nGiven, tan A\/2 – cot A\/2 + 2cot A
\n= {sin(A\/2)\/cos(A\/2)} – {cos(A\/2)\/sin(A\/2)} + 2cotA
\n= {sin\u00b2 (A\/2) – cos\u00b2 (A\/2)}\/{cos(A\/2) \u00d7 sin(A\/2)} + 2cotA
\n= -{cos\u00b2 (A\/2) – sin\u00b2 (A\/2)}\/{cos(A\/2) \u00d7 sin(A\/2)} + 2cotA
\n= -{cosA}\/{cos(A\/2) \u00d7 sin(A\/2)} + 2cotA (since cos\u00b2 A – sin\u00b2 A = cos\u00b2A )
\n= -{cos(2A\/2)}\/{cos(A\/2) \u00d7 sin(A\/2)} + 2cotA
\n= -{2 \u00d7 cosA}\/{2 \u00d7 cos(A\/2) \u00d7 sin(A\/2)} + 2cotA
\n= -{2cosA}\/{sin(2A\/2)} + 2cotA
\n= {-(2cosA)\/(sinA)} + 2cotA (since sin2A = 2 \u00d7 sinA \u00d7 cosA)
\n= -2cotA + 2cotA
\n= 0<\/p>\n<\/details>\n

\n

Question 3.
\nThe value of 4 \u00d7 sin x \u00d7 sin(x + \u03c0\/3) \u00d7 sin(x + 2\u03c0\/3) is
\n(a) sin x
\n(b) sin 2x
\n(c) sin 3x
\n(d) sin 4x<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) sin 3x
\nGiven, 4 \u00d7 sin x \u00d7 sin(x + \u03c0\/3) \u00d7 sin(x + 2\u03c0\/3)
\n= 4 \u00d7 sin x \u00d7 {sin x \u00d7 cos \u03c0\/3 + cos x \u00d7 sin \u03c0\/3} \u00d7 {sin x \u00d7 cos 2\u03c0\/3 + cos x \u00d7 sin 2\u03c0\/3}
\n= 4 \u00d7 sin x \u00d7 {(sin x)\/2 + (\u221a3 \u00d7 cos x)\/2} \u00d7 {-(sin x)\/2 + (\u221a3 \u00d7 cos x)\/2}
\n= 4 \u00d7 sin x \u00d7 {-(sin\u00b2 x)\/4 + (3 \u00d7 cos\u00b2 x)\/4}
\n= sin x \u00d7 {-sin\u00b2 x + 3 \u00d7 cos\u00b2 x}
\n= sin x \u00d7 {-sin\u00b2 x + 3 \u00d7 (1 – sin\u00b2 x)}
\n= sin x \u00d7 {-sin\u00b2 x + 3 – 3 \u00d7 sin\u00b2 x}
\n= sin x \u00d7 {3 – 4 \u00d7 sin\u00b2 x}
\n= 3 \u00d7 sin x – 4sin\u00b3 x
\n= sin 3x
\nSo, 4 \u00d7 sin x \u00d7 sin(x + \u03c0\/3) \u00d7 sin(x + 2\u03c0\/3) = sin 3x<\/p>\n<\/details>\n

\n

Question 4.
\nIf tan x = (cos 9 + sin 9)\/(cos 9 – sin 9), then x =
\n(a) 45
\n(b) 54
\n(c) 36
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 54
\nGiven, tan x = (cos 9 + sin 9)\/(cos 9 – sin 9)
\n\u21d2 tan x = {cos 9(1 + sin 9\/cos 9)}\/{cos 9(1 – sin 9\/cos 9)}
\n\u21d2 tan x = (1 + tan 9)}\/(1 – tan 9)
\n\u21d2 tan x = (tan 45 + tan 9)}\/(1 – tan 45 \u00d7 tan 9) {since tan 45 = 1}
\n\u21d2 tan x = tan(45 + 9) {Apply tan(A + B) formula}
\n\u21d2 tan x = tan(54)
\n\u21d2 x = 54<\/p>\n<\/details>\n

\n

Question 5.
\nIn a triangle ABC, sin A – cos B = cos C, then angle B is
\n(a) \u03c0\/2
\n(b) \u03c0\/3
\n(c) \u03c0\/4
\n(d) \u03c0\/6<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) \u03c0\/2
\nGiven, sin A – cos B = sin C
\n\u21d2 sin A = cos B + sin C
\n\u21d2 2 \u00d7 sin (A\/2) \u00d7 cos (A\/2) = 2 \u00d7 cos {(B + C)\/2} \u00d7 cos {(B – C)\/2}
\n\u21d2 2 \u00d7 sin (A\/2) \u00d7 cos (A\/2) = 2 \u00d7 cos {\u03c0\/2 – A\/2} \u00d7 cos {(B – C)\/2}
\n\u21d2 2 \u00d7 sin (A\/2) \u00d7 cos (A\/2) = 2 \u00d7 sin (A\/2) \u00d7 cos {(B – C)\/2}
\n\u21d2 cos (A\/2) = cos {(B – C)\/2}
\n\u21d2 A\/2 = (B – C)\/2
\n\u21d2 A = B – C
\n\u21d2 B = A + C
\n\u21d2 B = \u03c0 – B {Since A + B + C = \u03c0}
\n\u21d2 2B = \u03c0
\n\u21d2 B = \u03c0\/2<\/p>\n<\/details>\n

\n

Question 6.
\nThe value of cos 420\u00b0 is
\n(a) 0
\n(b) 1
\n(c) 1\/2
\n(d) \u221a3\/2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 1\/2
\ncos 420\u00b0 = cos(360\u00b0 + 60\u00b0 ) = cos 60\u00b0 = 1\/2<\/p>\n<\/details>\n

\n

Question 7.
\nIf in a triangle ABC, tan A + tan B + tan C = 6 then the value of cot A \u00d7 cot B \u00d7 cot C is
\n(a) 1\/2
\n(b) 1\/3
\n(c) 1\/4
\n(d) 1\/6<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 1\/6
\nGiven tanA + tanB + tanC = 6
\nNow tan(A + B + C) = {(tanA + tanB + tanC) – tanA \u00d7 tanB \u00d7 tanC}\/{1 – (tanA \u00d7 tanB + tanB \u00d7 tanC + tanA \u00d7 tanC)}
\nWe know that,
\nA + B + C = \u03c0
\n\u21d2 tan(A + B + C) = tan \u03c0
\n\u21d2 tan(A + B + C) = 0
\nNow
\n0 = {(tanA + tanB + tanC) – tanA \u00d7 tanB \u00d7 tanC}\/{1 – (tanA \u00d7 tanB + tanB \u00d7 tanC + tanA \u00d7 tanC)}
\n\u21d2 tanA + tanB + tanC – tanA \u00d7 tanB \u00d7 tanC = 0
\n\u21d2 tanA + tanB + tanC = tanA \u00d7 tanB \u00d7 tanC
\n\u21d2 tanA \u00d7 tanB \u00d7 tanC = 6
\n\u21d2 (1\/cotA) \u00d7 (1\/cotB) \u00d7 (1\/cotC) = 6
\n\u21d2 1\/(cot A \u00d7 cot B \u00d7 cot C) = 6
\n\u21d2 cot A \u00d7 cot B \u00d7 cot C = 1\/6<\/p>\n<\/details>\n

\n

Question 8.
\nIf a \u00d7 cos x + b \u00d7 cos x = c, then the value of (a \u00d7 sin x – b \u00d7 cos x)\u00b2 is
\n(a) a\u00b2 + b\u00b2 + c\u00b2
\n(b) a\u00b2 – b\u00b2 – c\u00b2
\n(c) a\u00b2 – b\u00b2 + c\u00b2
\n(d) a\u00b2 + b\u00b2 – c\u00b2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) a\u00b2 + b\u00b2 – c\u00b2
\nWe have
\n(a \u00d7 cos x + b \u00d7 sin x)\u00b2 + (a \u00d7 sin x – b \u00d7 cos x)\u00b2 = a\u00b2 + b\u00b2
\n\u21d2 c\u00b2 + (a \u00d7 sin x – b \u00d7 cos x)\u00b2 = a\u00b2 + b\u00b2
\n\u21d2 (a \u00d7 sin x – b \u00d7 cos x)\u00b2 = a\u00b2 + b\u00b2 – c\u00b2<\/p>\n<\/details>\n

\n

Question 9.
\nWhen the length of the shadow of a pole is equal to the height of the pole, then the elevation of source of light is
\n(a) 30\u00b0
\n(b) 60\u00b0
\n(c) 75\u00b0
\n(d) 45\u00b0<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 45\u00b0
\n\"MCQ
\nLet AB is the length of the pole and BC is the shadow of the pole.
\nGiven AB = BC
\nNow from triangle ABC,
\ntan \u03b8 = AB\/BC
\n\u21d2 tan \u03b8 = 1
\n\u21d2 \u03b8 = 45\u00b0
\nSo, the elevation of source of light is 45\u00b0<\/p>\n<\/details>\n

\n

Question 10.
\nIn any triangle ABC, if cos A\/a = cos B\/b = cos C\/c and the side a = 2, then the area of the triangle is
\n(a) \u221a3
\n(b) \u221a3\/4
\n(c) \u221a3\/2
\n(d) 1\/\u221a3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) \u221a3
\nGiven cos A\/a = cos B\/b = cos C\/c
\n= cos A\/k \u00d7 sin A = cos B\/k \u00d7 sin B = cos C\/k \u00d7 sin C {since sin A\/a = sin B\/b = sin C\/c = k}
\n= cot A = cot B = cot C
\n\u21d2 A = B = C = 60
\nSo, triangle ABS is equilateral.
\nNow area of the triangle = (\u221a3\/4) \u00d7 a\u00b2 = (\u221a3\/4) \u00d7 2\u00b2 = (\u221a3\/4) \u00d7 4 = \u221a3<\/p>\n<\/details>\n

\n

Question 11.
\nThe least values of cos\u00b2 \u03b8 + sec\u00b2 \u03b8 is
\n(a) 0
\n(b) 1
\n(c) 2
\n(d) more than 2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 2
\nIf a \u00d7 cos\u00b2 \u03b8 + b \u00d7 sec\u00b2 \u03b8 is given,
\nthen the least value = 2\u221aab
\nNow given, cos\u00b2 \u03b8 + sec\u00b2 \u03b8
\nHere, a = 1, b = 1
\nNow, least value = 2\u221a(1 \u00d7 1) = 2 \u00d7 1 = 2<\/p>\n<\/details>\n

\n

Question 12.
\nThe equation (cos p – 1) x\u00b2 + cos p \u00d7 x + sin p = 0, where x is a variable, has real roots. Then the interval of p may be any one of the following:
\n(a) (0, \u03c0)
\n(b) (\u2212\u03c0\/2, \u03c0\/2)
\n(c) (0, \u03c0)
\n(d) (\u2212\u03c0, 0)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) (0, \u03c0)
\nThe equation (cos p – 1)
\nx\u00b2 + cos p \u00d7 x + sin p = 0, where x is a variable, has real roots.
\nNow, for real roots,
\nDiscriminant \u2265 0
\n\u21d2 cos\u00b2 p – 4(cosp – 1)sinp \u2265 0
\n\u21d2 (cosp – 2sinp)\u00b2 – 4sin\u00b2 p + 4sinp \u2265 0
\n\u21d2 (cosp – 2sinp)\u00b2 + 4sin p(1 – sinp) \u2265 0 ………..1
\nNow, 1 – sinp \u2265 0
\n\u21d2 For all real p such that 0 < p < \u03c0
\nSo that 4sin p(1 – sinp) \u2265 0
\nSo, p \u2208 (0, \u03c0)<\/p>\n<\/details>\n

\n

Question 13.
\nThe value of (sec 8A – 1)\/(sec 4A – 1) is
\n(a) 0
\n(b) 1
\n(c) tan 8A\/tan 2A
\n(d) tan 2A\/tan 8A<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) tan 8A\/tan 2A
\nGiven, (sec 8A – 1)\/(sec 4A – 1)
\n= (1\/cos 8A – 1)\/(1\/cos 4A – 1)
\n= {(1 – cos 8A)\/cos 8A}\/{(1 – cos 4A)\/cos 4A}
\n= {(1 – cos 8A) \u00d7 cos 4A}\/{(1 – cos 4A) \u00d7 cos 8A}
\n= (2sin\u00b2 4A \u00d7 cos 4A}\/{2sin\u00b2 2A \u00d7 cos 8A} {since cos 2A = 1 – 2sin\u00b2 A}
\n= (2sin 4A \u00d7 sin 4A \u00d7 cos 4A}\/{2sin 2A \u00d7 sin 2A \u00d7 cos 8A}
\n= (sin 8A \u00d7 sin 4A}\/{2sin 2A \u00d7 sin 2A \u00d7 cos 8A} {since sin 2A = 2\u00d7sin A \u00d7 cos A}
\n= (sin 8A \u00d7 2sin 2A \u00d7 cos 2A}\/{2sin 2A \u00d7 sin 2A \u00d7 cos 8A}
\n= (sin 8A \u00d7 cos 2A}\/{sin 2A \u00d7 cos 8A}
\n= (sin 8A\/cos 8A)\/(sin 2A\/cos 2A)
\n= tan 8A\/tan 2A
\nSo, (sec 8A – 1)\/(sec 4A – 1) = tan 8A\/tan 2A<\/p>\n<\/details>\n

\n

Question 14.
\nThe value of (sin7x + sin5x) \/(cos7x + cos5x) + (sin9x + sin3x) \/ (cos9x + cos3x) is
\n(a) tan6x
\n(b) 2 tan6x
\n(c) 3 tan6x
\n(d) 4 tan6x<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 2 tan6x
\nGiven, (sin7x + sin5x) \/(cos7x + cos5x) + (sin9x + sin3x) \/ (cos9x + cos3x)
\n\u21d2 [{2\u00d7sin(7x + 5x)\/2 \u00d7 cos(7x – 5x)\/2}\/{2 \u00d7 cos(7x + 5x)\/2 \u00d7 cos(7x – 5x)\/2}] +
\n[{2\u00d7sin(9x + 3x)\/2 \u00d7 cos(9x – 3x)\/2}\/{2 \u00d7 cos(9x + 3x)\/2 \u00d7 cos(9x – 3x)\/2}]
\n\u21d2 [{2 \u00d7 sin6x \u00d7 cosx}\/{2 \u00d7 cos6x \u00d7 cosx}] + [{2 \u00d7 sin6x \u00d7 cosx}\/{2 \u00d7 cos6x \u00d7 cosx}]
\n\u21d2 (sin6x\/cos6x) + (sin6x\/cos6x)
\n\u21d2 tan6x + tan6x
\n\u21d2 2 tan6x<\/p>\n<\/details>\n

\n

Question 15.
\nIf x > 0 then the value of f(x) = -3 \u00d7 cos\u221a(3 + x + x\u00b2) lie in the interval
\n(a) [-1, 1]
\n(b) [-2, 2]
\n(c) [-3, 3]
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) [-3, 3]
\nGiven x > 0 then 3 + x + x\u00b2 > 0
\nNow, -1 \u2264 cos\u221a(3 + x + x\u00b2 ) \u2264 1 {Since -1 \u2264 cosx \u2264 1}
\n\u21d2 3 \u2265 -3 \u00d7 cos\u221a(3 + x + x\u00b2 ) \u2265 -3 {Multiply by -3}
\n\u21d2 -3 \u2264 f(x) \u2264 3
\n\u21d2 f(x) \u2208 [-3, 3]<\/p>\n<\/details>\n

\n

Question 16.
\nThe value of cos 4A – cos 4B is
\n(a) (cos A – cos B) \u00d7 (cos A + cos B) \u00d7 (cos A – sin B) \u00d7 (cos A + sin B)
\n(b) 2(cos A – cos B) \u00d7 (cos A + cos B) \u00d7 (cos A – sin B) \u00d7 (cos A + sin B)
\n(c) 4(cos A – cos B) \u00d7 (cos A + cos B) \u00d7 (cos A – sin B) \u00d7 (cos A + sin B)
\n(d) 8(cos A – cos B) \u00d7 (cos A + cos B) \u00d7 (cos A – sin B) \u00d7 (cos A + sin B)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 8(cos A – cos B) \u00d7 (cos A + cos B) \u00d7 (cos A – sin B) \u00d7 (cos A + sin B)
\nGiven, cos 4A – cos 4B
\n= 2cos\u00b2 2A – 1 – (2cos2 2B – 1) {since 2cos\u00b2 x – 1 = cos 2x}
\n= 2cos\u00b2 2A – 1 – 2cos\u00b2 2B + 1
\n= 2cos\u00b2 2A – 2cos\u00b2 2B
\n= 2(cos\u00b2 2A – cos\u00b2 2B)
\n= 2(cos 2A – cos 2B) \u00d7 (cos 2A + cos 2B)
\n= 2{2cos\u00b2 A – 1 – (2cos\u00b2 B – 1)} \u00d7 {2cos\u00b2 A – 1 + 1 – 2sin\u00b2 B} {since 1 – 2sin\u00b2 x = cos 2x}
\n= 2{2cos\u00b2 A – 1 – 2cos\u00b2 B + 1} \u00d7 {2cos\u00b2 A – 1 + 1 – 2sin\u00b2 B}
\n= 2{2cos\u00b2 A – 2cos\u00b2 B} \u00d7 {2cos\u00b2 A – 2sin\u00b2 B}
\n= 2 \u00d7 2 \u00d7 2{cos\u00b2 A – cos\u00b2 B} \u00d7 {cos\u00b2 A – sin\u00b2 B}
\n= 8(cos A – cos B) \u00d7 (cos A + cos B) \u00d7 (cos A – sin B) \u00d7 (cos A + sin B)
\nSo, cos 4A – cos 4B = 8(cos A – cos B) \u00d7 (cos A + cos B) \u00d7 (cos A – sin B) \u00d7 (cosA + sin B)<\/p>\n<\/details>\n

\n

Question 17.
\nThe value of cos 420\u00b0 is
\n(a) 0
\n(b) 1
\n(c) 1\/2
\n(d) \u221a3\/2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 1\/2
\ncos 420\u00b0 = cos(360\u00b0 + 60\u00b0 ) = cos 60\u00b0 = 1\/2<\/p>\n<\/details>\n

\n

Question 18.
\nIn a \u0394ABC, (b + c) cos A + (c + a) cos B + (a + b) cos C is equal to
\n(a) a + b + c
\n(b) 0
\n(c) none of these
\n(d) Rr<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) a + b + c
\nGiven (b + c) cos A + (c + a) cos B + (a + b) cos C
\n= b \u00d7 cos A + c \u00d7 cos A + c \u00d7 cos B + a \u00d7 cos B + a \u00d7 cos C + b \u00d7 cos C
\n= (b \u00d7 cos C + c \u00d7 cos B) + (c \u00d7 cos A + a \u00d7 cos C) + (b \u00d7 cos A + a \u00d7 cos B)
\n= a + b + c {since b \u00d7 cos C + c \u00d7 cos B = a, c \u00d7 cos A + a \u00d7 cos C = b, b \u00d7 cos A + a \u00d7 cos B = c}<\/p>\n<\/details>\n

\n

Question 19.
\ntan\u00b2 \u03b8 = 1 – a\u00b2 then the value of sec \u03b8 + tan\u00b3 \u03b8 \u00d7 cosec \u03b8 is
\n(a) (2 – a\u00b2)
\n(b) (2 – a\u00b2)1\/2<\/sup>
\n(c) (2 – a\u00b2)3\/2<\/sup>
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) (2 – a\u00b2)3\/2<\/sup>
\nGiven, tan\u00b2 \u03b8 = 1 – a\u00b2
\n\u21d2 tan \u03b8 = \u221a(1 – a\u00b2)
\n\"MCQ
\nFrom the figure and apply Pythagorus theorem,
\nAC\u00b2 = AB\u00b2 + BC\u00b2
\n\u21d2 AC\u00b2 = {\u221a(1 – a\u00b2)}\u00b2 + 12
\n\u21d2 AC\u00b2 = 1 – a\u00b2 + 1
\n\u21d2 AC\u00b2 = 2 – a\u00b2
\n\u21d2 AC = \u221a(2 – a\u00b2)
\nNow, sec \u03b8 = \u221a(2 – a\u00b2)
\ncosec \u03b8 = \u221a(2 – a\u00b2)\/\u221a(1 – a\u00b2)
\nand tan \u03b8 = \u221a(1 – a\u00b2)
\nGiven, sec \u03b8 + tan\u00b3 \u03b8 \u00d7 cosec \u03b8
\n= \u221a(2 – a\u00b2) + {(1 – a\u00b2)3\/2<\/sup> \u00d7 \u221a(2 – a\u00b2)\/\u221a(1 – a\u00b2)}
\n= \u221a(2 – a\u00b2) + {(1 – a\u00b2) \u00d7 (1 – a\u00b2) \u00d7 \u221a(2 – a\u00b2)\/\u221a(1 – a\u00b2)}
\n= \u221a(2 – a\u00b2) + (1 – a\u00b2) \u00d7 \u221a(2 – a\u00b2)
\n= \u221a(2 – a\u00b2) \u00d7 (1 + 1 – a\u00b2)
\n= \u221a(2 – a\u00b2) \u00d7 (2 – a\u00b2)
\n= (2 – a\u00b2)3\/2
\nSo, sec \u03b8 + tan\u00b3 \u03b8 \u00d7 cosec \u03b8 = (2 – a\u00b2)3\/2<\/sup><\/p>\n<\/details>\n

\n

Question 20.
\nThe value of cos(\u03c0\/7) \u00d7 cos(2\u03c0\/7) \u00d7 cos(4\u03c0\/7) is
\n(a) -1\/2
\n(b) -1\/4
\n(c) -1\/6
\n(d) -1\/8<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) -1\/8
\nWe know that cos A \u00d7 cos 2A \u00d7 cos 2\u00b2 A \u00d7 ……………… \u00d7 cos 2n-1<\/sup> A = sin (2\u207f A)\/{2\u207f \u00d7 sin A} ……………1
\nGiven, cos(\u03c0\/7) \u00d7 cos(2\u03c0\/7) \u00d7 cos(4\u03c0\/7)
\n= cos(\u03c0\/7) \u00d7 cos(2\u03c0\/7) \u00d7 cos(2\u00b2 \u03c0\/7)
\n= [sin (2\u00b3 \u00d7 \u03c0\/7) ]\/{2\u00b3 \u00d7 sin (\u03c0\/7)} ……………..from equation 1
\n= [sin (8\u03c0\/7) ]\/{8 \u00d7 sin (\u03c0\/7)}
\n= [sin (\u03c0 + \u03c0\/7) ]\/{8 \u00d7 sin (\u03c0\/7)}
\n= -sin (\u03c0\/7)\/{8 \u00d7 sin (\u03c0\/7)}
\n= -1\/8
\nSo, cos(\u03c0\/7) \u00d7 cos(2\u03c0\/7) \u00d7 cos(4\u03c0\/7) = -1\/8<\/p>\n<\/details>\n

\n

We believe the knowledge shared regarding NCERT MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers Pdf free download has been useful to the possible extent. If you have any other queries regarding CBSE Class 11 Maths Trigonometric Functions MCQs Multiple Choice Questions with Answers, feel free to reach us via the comment section and we will guide you with the possible solution.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can …<\/p>\n

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