JEE Mathematics Trigonometric Functions MCQs Set 19

Answer: (d) \(2n\pi+\frac{7\pi}4\)

Explanation:
1. \(\cos\theta>0\) and \(\tan\theta<0\) puts \(\theta\) in quadrant IV: \(\theta=\frac{7\pi}4\).
2. General value \(2n\pi+\frac{7\pi}4\).

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) \(\left(\frac{5\pi}4,\frac{7\pi}4\right)\)

Explanation:
1. \(2-\sin^2\theta+\sin\theta=0\) gives \(\sin\theta=-1\), \(\theta=\frac{3\pi}2\).
2. \(\frac{3\pi}2\) lies in \(\left(\frac{5\pi}4,\frac{7\pi}4\right)\).

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) \(4\)

Explanation:
1. \((2\sin x-1)(\sin x+3)=0\) gives \(\sin x=\frac12\).
2. In \([0,3\pi]\): \(\frac\pi6,\frac{5\pi}6,\frac{13\pi}6,\frac{17\pi}6\), so 4 values.

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) \(\cos7^\circ\)

Explanation:
1. \((\sin47^\circ-\sin25^\circ)+(\sin61^\circ-\sin11^\circ)=2\cos36^\circ(\sin11^\circ+\sin25^\circ)=4\cos36^\circ\sin18^\circ\cos7^\circ\).
2. \(4\sin18^\circ\cos36^\circ=1\), so the value is \(\cos7^\circ\).

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) \(-\sqrt2\) and \(\sqrt2\) both inclusive

Explanation:
1. \(\sin\theta+\cos\theta=\sqrt2\sin\left(\theta+\frac\pi4\right)\).

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) \(\sin2x=\dfrac4{(2n+1)\pi}\)

Explanation:
1. \(\cot x=\frac\pi2-\tan x+n\pi\), so \(\tan x+\cot x=(2n+1)\frac\pi2\).
2. \(\dfrac2{\sin2x}=(2n+1)\frac\pi2\), hence \(\sin2x=\dfrac4{(2n+1)\pi}\).

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) 110.9 cm

Explanation:
1. \(\theta=\frac{31}{60}\times\frac\pi{180}=\frac{31\times22}{60\times180\times7}=0.00902\) rad.
2. Distance \(=\dfrac1\theta=110.9\) cm.

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) 7.92 km

Explanation:
1. Revolutions in 180 s \(=3600\); each is \(2\pi\times35=220\) cm.
2. Distance \(=792000\) cm \(=7.92\) km.

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) 22 : 13

Explanation:
1. \(r_1\cdot65=r_2\cdot110\), so \(r_1:r_2=110:65=22:13\).

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) \(194\)

Explanation:
1. \(\tan^2A+\cot^2A=16-2=14\).
2. \(\tan^4A+\cot^4A=196-2=194\).

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) \(\pm\sqrt{\dfrac{1-n^2}{m^2-1}}\)

Explanation:
1. \(m^2\sin^2B+n^2\cos^2B=1\) gives \(\sin^2B=\dfrac{1-n^2}{m^2-n^2}\), \(\cos^2B=\dfrac{m^2-1}{m^2-n^2}\).
2. \(\tan^2B=\dfrac{1-n^2}{m^2-1}\).

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: (c) \(\sin3\alpha\)

Explanation:
1. \(\sin\left(\alpha+\frac{2\pi}3\right)=\sin\left(\frac\pi3-\alpha\right)\).
2. \(4\sin\alpha\sin\left(\frac\pi3-\alpha\right)\sin\left(\frac\pi3+\alpha\right)=\sin3\alpha\).

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) \(\frac\pi4\)

Explanation:
1. With \(u=\sin x+\cos x\): \(u^{u^2}=2\), satisfied by \(u=\sqrt2\) since \((\sqrt2)^2=2\).
2. \(\sqrt2\sin\left(x+\frac\pi4\right)=\sqrt2\) gives \(x=\frac\pi4\).

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) \(\dfrac{p^2+1}{2p}\)

Explanation:
1. \(\sec\theta-\tan\theta=\frac1p\). Adding: \(2\sec\theta=p+\frac1p\), so \(\sec\theta=\dfrac{p^2+1}{2p}\).

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. \(\sqrt3\cos\theta-\sin\theta=1\) gives \(\cos\left(\theta+\frac\pi6\right)=\frac12\), so \(\theta=\frac\pi6\) or \(\frac{3\pi}2\).
2. At \(\theta=\frac{3\pi}2\), \(\cos\theta=0\) (not defined), so only \(\theta=\frac\pi6\).

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) \(n\pi+(-1)^n\frac\pi4-\frac\pi4\)

Explanation:
1. \(\sin\left(\theta+\frac\pi4\right)=\frac1{\sqrt2}\), so \(\theta+\frac\pi4=n\pi+(-1)^n\frac\pi4\).
2. This covers both \(2m\pi\) (n even) and \(2m\pi+\frac\pi2\) (n odd).

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) \(\frac{n\pi}4\) or \(n\pi\pm\frac\pi3\)

Explanation:
1. \(2\sin4\theta\cos2\theta+\sin4\theta=\sin4\theta(2\cos2\theta+1)=0\).
2. \(\sin4\theta=0\): \(\theta=\frac{n\pi}4\). \(\cos2\theta=-\frac12\): \(\theta=n\pi\pm\frac\pi3\).

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) \(2n\pi-\frac\pi4\)

Explanation:
1. \(\cos\theta-\sin\theta=\sqrt2\) gives \(\cos\left(\theta+\frac\pi4\right)=1\).
2. \(\theta=2n\pi-\frac\pi4\).

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) \(n\pi\pm\frac\pi6\)

Explanation:
1. \(\cos^2\theta=\frac34\), so \(\cos\theta=\pm\frac{\sqrt3}2\) and \(\theta=n\pi\pm\frac\pi6\).

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) \(\theta=\frac{n\pi}7+\frac\pi{14}\)

Explanation:
1. \(\tan5\theta=\tan\left(\frac\pi2-2\theta\right)\) gives \(5\theta=n\pi+\frac\pi2-2\theta\).
2. \(7\theta=n\pi+\frac\pi2\), so \(\theta=\frac{n\pi}7+\frac\pi{14}\).

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) \(\theta=n\pi+\frac\pi4\)

Explanation:
1. \(\dfrac{\tan(\theta+15^\circ)}{\tan(\theta-15^\circ)}=3\); componendo and dividendo gives \(\dfrac{\sin2\theta}{\sin30^\circ}=\dfrac{3+1}{3-1}=2\).
2. \(\sin2\theta=1\), so \(\theta=n\pi+\frac\pi4\).

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) \(\dfrac{m+n}{m-n}\)

Explanation:
1. Write \(m\sin(T-\alpha)=n\sin(T+\alpha)\) with \(T=\theta+\alpha\) and expand.
2. \((m-n)\sin T\cos\alpha=(m+n)\cos T\sin\alpha\), so \(\tan T\cot\alpha=\dfrac{m+n}{m-n}\).

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) \(\frac16\)

Explanation:
1. Divide by \(\cos\theta\): \(\dfrac{5\tan\theta-3}{5\tan\theta+2}=\dfrac{4-3}{4+2}=\dfrac16\).

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) \(\tan\frac A2\)

Explanation:
1. Numerator \(=2\sin\frac A2\left(\cos\frac A2+\sin\frac A2\right)\) and denominator \(=2\cos\frac A2\left(\sin\frac A2+\cos\frac A2\right)\).
2. The ratio is \(\tan\frac A2\).

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) \(\cot\alpha\)

Explanation:
1. \(2\cos\frac{x+y}2\cos\frac{x-y}2=-\cos\alpha\) and \(2\sin\frac{x+y}2\cos\frac{x-y}2=-\sin\alpha\).
2. Dividing: \(\tan\frac{x+y}2=\tan\alpha\), so \(\cot\frac{x+y}2=\cot\alpha\).

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.