JEE Mathematics Inverse Trigonometric Functions MCQs Set 17

Answer: (d) \( -\frac{\pi}{10} \)

Explanation:
1. \( \cos\frac{33\pi}{5}=\cos\left(6\pi+\frac{3\pi}{5}\right)=\cos\frac{3\pi}{5}=\sin\left(\frac\pi2-\frac{3\pi}5\right)=\sin\left(-\frac{\pi}{10}\right) \).
2. \( -\frac\pi{10} \) lies in \( \left[-\frac\pi2,\frac\pi2\right] \), so the principal value is \( -\frac\pi{10} \).

Teacher's Note:
1. Reduce the angle to the principal range first.

Answer: (a) \( \tan1>\tan^{-1}1 \)

Explanation:
1. \( \tan 1 \) (1 radian) \( \approx1.557 \).
2. \( \tan^{-1}1=\frac\pi4\approx0.785 \), so \( \tan1>\tan^{-1}1 \).

Teacher's Note:
1. Remember that 1 is in radians in \( \tan1 \).

Answer: (d) \( -1\le x<\frac{1}{\sqrt2} \)

Explanation:
1. Since \( \cos^{-1}x=\frac\pi2-\sin^{-1}x \), the inequality becomes \( \frac\pi2>2\sin^{-1}x \), i.e. \( \sin^{-1}x<\frac\pi4 \).
2. So \( x<\sin\frac\pi4=\frac1{\sqrt2} \), and with the domain \( x\ge-1 \) we get \( -1\le x<\frac1{\sqrt2} \).

Teacher's Note:
1. Use \( \sin^{-1}x+\cos^{-1}x=\frac\pi2 \) and keep the domain in mind.

Answer: (c) \( 3\pi-10 \)

Explanation:
1. \( 10 \) radians lies between \( 3\pi\approx9.42 \) and \( \frac{7\pi}{2}\approx11 \), outside \( \left[-\frac\pi2,\frac\pi2\right] \).
2. \( \sin10=\sin(3\pi-10) \) and \( 3\pi-10\approx-0.58 \) is in the principal range, so the value is \( 3\pi-10 \).

Teacher's Note:
1. Find an equivalent angle in \( [-\frac\pi2,\frac\pi2] \).

Answer: (b) 1

Explanation:
1. \( \cos^{-1}\frac35=\sin^{-1}\frac45 \) (both are the angle of the 3-4-5 triangle).
2. So the left side is 0, giving \( \cos^{-1}x=0 \) and \( x=1 \).

Teacher's Note:
1. Both inverse functions describe the same angle.

Answer: (a) 20

Explanation:
1. Each \( \sin^{-1}x_i\le\frac\pi2 \), so 20 terms give at most \( 10\pi \).
2. Equality needs every \( \sin^{-1}x_i=\frac\pi2 \), so each \( x_i=1 \) and the sum is 20.

Teacher's Note:
1. Use the maximum value of \( \sin^{-1}x \).

Answer: (a) \( [1,2] \)

Explanation:
1. We need \( 0\le x-1\le1 \) for the square root and the sine inverse.
2. So \( 1\le x\le2 \).

Teacher's Note:
1. The argument of \( \sin^{-1} \) must be in \( [-1,1] \).

Answer: (a) \( [0,1] \)

Explanation:
1. We need \( -1\le2x-1\le1 \), that is \( 0\le2x\le2 \).
2. So \( x\in[0,1] \).

Teacher's Note:
1. Solve \( -1\le2x-1\le1 \).

Answer: (b) 1

Explanation:
1. Using \( \cot^{-1}x=\frac\pi2-\tan^{-1}x \): \( 2\tan^{-1}x+\frac\pi2=\pi \).
2. So \( \tan^{-1}x=\frac\pi4 \) and \( x=1 \).

Teacher's Note:
1. Replace \( \cot^{-1}x \) by \( \frac\pi2-\tan^{-1}x \).

Answer: (c) \( \frac12 \)

Explanation:
1. \( \sin^{-1}\left(-\frac{\sqrt3}2\right)=-\frac\pi3 \).
2. The expression is \( \sin\left(\frac\pi2+\frac\pi3\right)=\cos\frac\pi3=\frac12 \).

Teacher's Note:
1. Evaluate the inner function first.

Answer: (a) \( \frac{2}{3\sqrt5} \)

Explanation:
1. Let \( \theta=\cos^{-1}\left(-\frac27\right) \), so \( \cos\theta=-\frac27 \) and \( \sin\theta=\frac{\sqrt{45}}{7}=\frac{3\sqrt5}{7} \).
2. \( \tan\left(\theta-\frac\pi2\right)=-\cot\theta=-\frac{-2/7}{3\sqrt5/7}=\frac{2}{3\sqrt5} \).

Teacher's Note:
1. Use \( \tan(\theta-\frac\pi2)=-\cot\theta \).

Answer: (b) \( \frac{5\pi}{6} \)

Explanation:
1. \( \sec^{-1}2=\frac\pi3 \), so \( 2\sec^{-1}2=\frac{2\pi}3 \).
2. \( \sin^{-1}\frac12=\frac\pi6 \), so the sum is \( \frac{2\pi}{3}+\frac\pi6=\frac{5\pi}6 \).

Teacher's Note:
1. Evaluate each principal value.

Answer: (a) \( \frac\pi5 \)

Explanation:
1. \( \cot^{-1}x+\cot^{-1}y=\left(\frac\pi2-\tan^{-1}x\right)+\left(\frac\pi2-\tan^{-1}y\right)=\pi-\frac{4\pi}5 \).
2. So the value is \( \frac\pi5 \).

Teacher's Note:
1. Use \( \cot^{-1}t=\frac\pi2-\tan^{-1}t \).

Answer: (d) \( \frac7{24} \)

Explanation:
1. Let \( \cos\theta=\frac7{25} \); then \( \sin\theta=\frac{24}{25} \).
2. \( \cot\theta=\frac{7}{24} \).

Teacher's Note:
1. Draw a 7-24-25 triangle.

Answer: (c) 0

Explanation:
1. \( \tan\frac{5\pi}{6}=-\frac1{\sqrt3} \), so \( \tan^{-1}\left(\tan\frac{5\pi}6\right)=-\frac\pi6 \).
2. \( \cos\frac{13\pi}{6}=\cos\frac\pi6 \), so \( \cos^{-1}(\ldots)=\frac\pi6 \). The sum is 0.

Teacher's Note:
1. Bring each angle into the principal range.

Answer: (b) \( -1 \)

Explanation:
1. \( \cos^{-1}\left(-\frac{\sqrt3}2\right)=\frac{5\pi}6 \).
2. \( \cos\left(\frac{5\pi}6+\frac\pi6\right)=\cos\pi=-1 \).

Teacher's Note:
1. Evaluate the inverse cosine first.

Answer: (c) \( -\frac\pi{12} \)

Explanation:
1. \( \tan^{-1}\left(-\frac1{\sqrt3}\right)=-\frac\pi6 \), \( \cot^{-1}\frac1{\sqrt3}=\frac\pi3 \) and \( \tan^{-1}(-1)=-\frac\pi4 \).
2. Sum \( =-\frac\pi6+\frac\pi3-\frac\pi4=-\frac\pi{12} \).

Teacher's Note:
1. Use the principal ranges: \( \cot^{-1}\in(0,\pi) \).

Answer: (a) \( \tan^{-1}\frac43-x \)

Explanation:
1. Let \( \cos\alpha=\frac35 \), \( \sin\alpha=\frac45 \) with \( \alpha=\tan^{-1}\frac43\approx0.927 \). The expression is \( \cos^{-1}\cos(x-\alpha) \).
2. For \( x\in\left[-\frac{3\pi}{4},\frac\pi4\right] \), \( x-\alpha\in[-\pi,\ \pi/4-0.927] \) is negative, so the principal value is \( \alpha-x \) (which lies in \( [0,\pi] \)).
3. So it is \( \tan^{-1}\frac43-x \).

Teacher's Note:
1. The principal value of \( \cos^{-1}\cos t \) is \( |t| \) for \( t\in[-\pi,\pi] \).

Answer: (c) \( \frac32 \)

Explanation:
1. The second condition gives \( x=y \), and then \( 2\cos^{-1}x=\frac\pi2 \), so \( x=y=\cos\frac\pi4=\frac{1}{\sqrt2} \).
2. \( x^2+xy+y^2=\frac12+\frac12+\frac12=\frac32 \).

Teacher's Note:
1. Equal tangents mean equal numbers.

Answer: (a) \( \frac16 \)

Explanation:
1. \( \frac{5x}{1-6x^2}=1 \) gives \( 6x^2+5x-1=0 \), so \( x=\frac16 \) or \( x=-1 \).
2. For \( x=-1 \) both inverse tangents are negative and cannot add up to \( \frac\pi4 \), so only \( x=\frac16 \) works.

Teacher's Note:
1. Reject roots that do not satisfy the original equation.

Answer: (c) 0.96

Explanation:
1. \( \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta} \) with \( \tan\theta=0.75 \).
2. \( =\frac{1.5}{1.5625}=0.96 \).

Teacher's Note:
1. Use the formula for \( \sin2\theta \) in terms of \( \tan\theta \).

Answer: (d) \( \frac{2a}{1-a^2} \)

Explanation:
1. For \( a\in(0,1) \) both terms on the left equal \( 2\tan^{-1}a \), so the left side is \( 4\tan^{-1}a \).
2. The right side is \( 2\tan^{-1}x \), so \( \tan^{-1}x=2\tan^{-1}a \).
3. Therefore \( x=\frac{2a}{1-a^2} \).

Teacher's Note:
1. Convert each term to \( 2\tan^{-1}a \).

Answer: (d) \( \frac3{29} \)

Explanation:
1. If \( \cos\alpha=\frac1{5\sqrt2} \) then \( \tan\alpha=\frac{\sqrt{49}}{1}=7 \). If \( \sin\beta=\frac4{\sqrt{17}} \) then \( \tan\beta=4 \).
2. \( \tan(\alpha-\beta)=\frac{7-4}{1+28}=\frac3{29} \).

Teacher's Note:
1. Convert both angles to tangents.

Answer: (a) \( \sqrt{ab} \)

Explanation:
1. For the sum to be \( \frac\pi2 \) the product of the two tangents must be 1: \( \frac{ab}{x^2}=1 \).
2. So \( x=\sqrt{ab} \) (for positive a, b).

Teacher's Note:
1. The tangents of complementary angles are reciprocals.

Answer: (b) \( \sqrt5-2 \)

Explanation:
1. Let \( \theta=\frac12\cos^{-1}\frac2{\sqrt5} \), so \( \cos2\theta=\frac2{\sqrt5} \).
2. \( \tan^2\theta=\frac{1-\cos2\theta}{1+\cos2\theta}=\frac{\sqrt5-2}{\sqrt5+2}=(\sqrt5-2)^2 \).
3. So \( \tan\theta=\sqrt5-2 \).

Teacher's Note:
1. Use the half-angle formula for tangent.

Answer: (d) x

Explanation:
1. Write \( \frac{3\sin2x}{5+3\cos2x}=\frac{3\cdot2t/(1+t^2)}{5+3(1-t^2)/(1+t^2)}=\frac{6t}{8+2t^2}=\frac{3t}{4+t^2} \) with \( t=\tan x \).
2. Then \( \tan^{-1}\frac t4+\tan^{-1}\frac{3t}{4+t^2}=\tan^{-1}\frac{\frac t4+\frac{3t}{4+t^2}}{1-\frac{3t^2}{4(4+t^2)}}=\tan^{-1}t=x \).

Teacher's Note:
1. Express everything in \( t=\tan x \).

Answer: (a) \( x=3 \)

Explanation:
1. \( \sin^{-1}\frac1{\sqrt5}=\tan^{-1}\frac12 \), so \( \cot^{-1}x=\frac\pi4-\tan^{-1}\frac12=\tan^{-1}\frac{1-1/2}{1+1/2}=\tan^{-1}\frac13 \).
2. So \( \cot^{-1}x=\tan^{-1}\frac13=\cot^{-1}3 \), and \( x=3 \).

Teacher's Note:
1. Convert \( \sin^{-1}\frac1{\sqrt5} \) to an inverse tangent.

Answer: (d) \( -\frac{2\sqrt6}5 \)

Explanation:
1. \( 2\cos^{-1}x+\sin^{-1}x=\cos^{-1}x+\frac\pi2 \).
2. \( \cos\left(\cos^{-1}x+\frac\pi2\right)=-\sin(\cos^{-1}x)=-\sqrt{1-x^2}=-\sqrt{\frac{24}{25}}=-\frac{2\sqrt6}5 \).

Teacher's Note:
1. Use \( \sin^{-1}x=\frac\pi2-\cos^{-1}x \).

Answer: (c) \( \frac\pi4 \)

Explanation:
1. The general term is \( \cot^{-1}(2n^2)=\tan^{-1}\frac1{2n^2}=\tan^{-1}\frac1{2n-1}-\tan^{-1}\frac1{2n+1} \).
2. The series telescopes to \( \tan^{-1}1-\lim\tan^{-1}\frac1{2n+1}=\frac\pi4 \).

Teacher's Note:
1. Write \( \frac1{2n^2} \) as a difference of two arctangents.