JEE Mathematics Trigonometric Functions MCQs Set 18

Answer: (a) Only II is correct

Explanation:
1. \(\cos(x+y)-\cos(x-y)=-2\sin x\sin y\) so II is correct.
2. \(2\cos x\cos y=\cos(x+y)+\cos(x-y)\), so I is wrong.

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) A-4, B-1, C-2, D-3

Explanation:
1. \(25^\circ=\frac{5\pi}{36}\), \(-47.5^\circ=-\frac{19\pi}{72}\), \(240^\circ=\frac{4\pi}3\), \(520^\circ=\frac{26\pi}9\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) A-3, B-4, C-1, D-2

Explanation:
1. \(\frac{11}{16}\times\frac{180\times7}{22}=39.375^\circ=39^\circ22'30''\) and \(-4\times\frac{180\times7}{22}=-229.09^\circ=-229^\circ5'27''\).
2. \(\frac{5\pi}3=300^\circ\), \(\frac{7\pi}6=210^\circ\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) A-1, B-4, C-2, D-1, E-3, F-2

Explanation:
1. \(\cos(\pi-x)=-\cos x\), \(\sin(\pi-x)=\sin x\), \(\sin(\pi+x)=-\sin x\), \(\cos(\pi+x)=-\cos x\), \(\cos(2\pi-x)=\cos x\), \(\sin(2\pi-x)=-\sin x\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) A-2, B-1, C-4, D-3

Explanation:
1. \(1\) rad \(\approx57^\circ16'\), \(1^\circ=0.01746\) rad, \(3.75^\circ=\frac\pi{48}\), \(50.625^\circ=\frac{9\pi}{32}\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) A-3, B-5, C-1, D-2, E-4

Explanation:
1. A: \(\frac{-2\sin7x\sin2x}{2\cos10x\sin7x}=-\frac{\sin2x}{\cos10x}\). B: \(\tan4x\). C: \(\tan2x\). D: \(\tan\frac{x-y}2\). E: \(2\sin x\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (b) \(2\)

Explanation:
1. \(1410^\circ=4\times360^\circ-30^\circ\), so \(\sin(-1410^\circ)=-\sin(1410^\circ)=\sin30^\circ=\frac12\).
2. \(\mathrm{cosec}(-1410^\circ)=2\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) \(0\)

Explanation:
1. \(\cos\frac{10\pi}{13}=-\cos\frac{3\pi}{13}\) and \(\cos\frac{8\pi}{13}=-\cos\frac{5\pi}{13}\), so the sum is 0.

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (d) \(0\)

Explanation:
1. Squaring: \(1+2\sin\theta\cos\theta=1\), so \(\sin\theta\cos\theta=0\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) \(4\)

Explanation:
1. \(\cos A=\frac35\), \(\sin A=-\frac45\); \(\cos B=\frac45\), \(\sin B=-\frac35\).
2. \(2\sin A+4\sin B=-\frac85-\frac{12}5=-4\), so \(a=4\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (b) \(3\)

Explanation:
1. \(\frac{19\pi}3=6\pi+\frac\pi3\), so \(\tan=\sqrt3\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (d) \(2\)

Explanation:
1. \(\sin\left(-\frac{11\pi}3\right)=-\sin\frac{11\pi}3=\sin\frac\pi3=\frac{\sqrt3}2\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) \(3\)

Explanation:
1. \(\left(1+\frac{\sqrt3}2\right)\left(1-\frac{\sqrt3}2\right)=\frac14\) and \(\left(1+\frac12\right)\left(1-\frac12\right)=\frac34\).
2. Product \(=\frac3{16}\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) \(2\)

Explanation:
1. \(\cos x=-\frac15\), so \(\cot x=\dfrac{-1/5}{-2\sqrt6/5}=\dfrac1{2\sqrt6}\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (d) \(1\)

Explanation:
1. \(3\cdot\frac12\cdot2-4\cdot\frac12\cdot1=3-2=1\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) \(3\)

Explanation:
1. \(2\cdot\frac14+4\cdot\frac14=\frac32\).
2. \(\frac m{m-1}=\frac32\) gives \(m=3\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) \(1\)

Explanation:
1. \(\cos\left(\frac{3\pi}2+x\right)=\sin x\) and \(\cot\left(\frac{3\pi}2-x\right)=\tan x\).
2. Expression \(=\sin x\cos x(\tan x+\cot x)=1\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) \(1\)

Explanation:
1. \(\cot3x=\dfrac{\cot x\cot2x-1}{\cot x+\cot2x}\), so \(\cot3x(\cot x+\cot2x)=\cot x\cot2x-1\).
2. Hence the expression equals 1.

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (b) Assertion is false, Reason is true

Explanation:
1. Equal arcs: \(r_1\cdot30^\circ=r_2\cdot70^\circ\) gives \(r_1:r_2=7:3\), not 21:10.
2. The reason is the correct definition of radian measure.

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (b) Assertion is false, Reason is true

Explanation:
1. The condition gives \(\sin\theta+\cos\theta=2k+1\) for an integer \(k\); within \([-\sqrt2,\sqrt2]\) only \(\pm1\) is possible.
2. So the assertion (\(\pm\sqrt2\)) is false while the reason is true.

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) Assertion is false, Reason is true

Explanation:
1. \(-2\times\frac{180}\pi\approx-114.59^\circ=-114^\circ35'\), not \(-114^\circ19'\).
2. The formula in the reason is correct.

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (d) Both are true and Reason explains Assertion

Explanation:
1. \(-\cot\left(x+\frac\pi3\right)=\tan\left(x+\frac\pi3+\frac\pi2\right)=\tan\left(x+\frac{5\pi}6\right)\).
2. Then \(2x=n\pi+x+\frac{5\pi}6\), so \(x=n\pi+\frac{5\pi}6\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (d) Assertion is false, Reason is true

Explanation:
1. The second hand makes one full turn, 360 degrees, in a minute, so the assertion is false.
2. The sexagesimal system does use degrees, so the reason is true.

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) \(0\)

Explanation:
1. \(\tan70^\circ=\cot20^\circ\) and \(\tan50^\circ=\cot40^\circ\).
2. \(\tan20^\circ-\cot20^\circ=-2\cot40^\circ\), which cancels \(2\tan50^\circ\), giving 0.

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (b) \(\frac{56}{65}\)

Explanation:
1. \(\sin(\alpha+\beta)=\frac5{13}\) and \(\cos(\alpha-\beta)=\frac45\).
2. \(\sin2\alpha=\frac5{13}\cdot\frac45+\frac{12}{13}\cdot\frac35=\frac{56}{65}\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.