JEE Mathematics Inverse Trigonometric Functions MCQs Set 15

Answer: (d) \( \sin^{-1}\left(\dfrac9{5\sqrt{10}}\right) \)

Explanation:
1. \( \alpha=\tan^{-1}\dfrac43 \), so \( \alpha-\beta=\tan^{-1}\dfrac{\frac43-\frac13}{1+\frac49}=\tan^{-1}\dfrac9{13} \).
2. For \( \tan\theta=\dfrac9{13} \) the hypotenuse is \( \sqrt{250}=5\sqrt{10} \), so \( \theta=\sin^{-1}\dfrac9{5\sqrt{10}} \). Answer: (d)

Teacher's Note:
1. Convert between inverse functions using a right triangle.

Answer: (c) \( 4\sin^2\alpha \)

Explanation:
1. Taking cosine: \( \dfrac{xy}2+\sqrt{1-x^2}\sqrt{1-\frac{y^2}4}=\cos\alpha \), so \( \sqrt{(1-x^2)\left(1-\frac{y^2}4\right)}=\cos\alpha-\dfrac{xy}2 \).
2. Squaring: \( 1-x^2-\dfrac{y^2}4+\dfrac{x^2y^2}4=\cos^2\alpha-xy\cos\alpha+\dfrac{x^2y^2}4 \), so \( x^2+\dfrac{y^2}4-xy\cos\alpha=\sin^2\alpha \) and \( 4x^2-4xy\cos\alpha+y^2=4\sin^2\alpha \). Answer: (c)

Teacher's Note:
1. Take the cosine of both sides and square.

Answer: (b) \( \dfrac\pi2-\sin^{-1}\left(\dfrac{56}{65}\right) \)

Explanation:
1. Difference \( =\sin^{-1}\left(\dfrac{12}{13}\cdot\dfrac45-\dfrac35\cdot\dfrac5{13}\right)=\sin^{-1}\dfrac{33}{65} \).
2. \( \sin^{-1}\dfrac{33}{65}=\dfrac\pi2-\cos^{-1}\dfrac{33}{65}=\dfrac\pi2-\sin^{-1}\dfrac{56}{65} \). Answer: (b)

Teacher's Note:
1. Use the formula for \( \sin^{-1}x-\sin^{-1}y \).

Answer: (c) \( \dfrac{\sqrt{145}}{12} \)

Explanation:
1. \( \cos^{-1}a+\cos^{-1}b=\dfrac\pi2 \) means \( \cos^{-1}b=\sin^{-1}a \), so \( a^2+b^2=1 \).
2. \( \dfrac4{9x^2}+\dfrac9{16x^2}=1 \) gives \( x^2=\dfrac{145}{144} \), and \( x\gt0 \) gives \( x=\dfrac{\sqrt{145}}{12} \). Answer: (c)

Teacher's Note:
1. If \( \cos^{-1}a+\cos^{-1}b=\pi/2 \) then \( a^2+b^2=1 \).

Answer: (d) \( \pi \)

Explanation:
1. \( 10\in\left[\dfrac{5\pi}2,\dfrac{7\pi}2\right] \), so \( x=3\pi-10 \).
2. \( 10\in[3\pi,4\pi] \), so \( y=4\pi-10 \). Hence \( y-x=\pi \). Answer: (d)

Teacher's Note:
1. Reduce \( 10 \) rad into the principal ranges.

Answer: (b) \( \dfrac{21}{19} \)

Explanation:
1. \( 1+\sum_{p=1}^n2p=1+n(n+1) \), and \( \cot^{-1}(1+n(n+1))=\tan^{-1}\dfrac{(n+1)-n}{1+n(n+1)}=\tan^{-1}(n+1)-\tan^{-1}n \).
2. The sum telescopes to \( \tan^{-1}20-\tan^{-1}1 \), so \( \cot(A-B)=\dfrac{1+20\cdot1}{20-1}=\dfrac{21}{19} \). Answer: (b)

Teacher's Note:
1. Telescope using \( \tan^{-1}x-\tan^{-1}y \).

Answer: (c) \( (\cot2,\infty) \)

Explanation:
1. \( (\cot^{-1}x-2)(\cot^{-1}x-5)\gt0 \), so \( \cot^{-1}x\lt2 \) or \( \cot^{-1}x\gt5 \). Since \( \cot^{-1}x\in(0,\pi) \) and \( 5\gt\pi \), only \( \cot^{-1}x\lt2 \) is possible.
2. \( \cot \) is decreasing, so \( x\gt\cot2 \). Answer: (c)

Teacher's Note:
1. Factor, then use the range \( (0,\pi) \).

Answer: (b) is a singleton

Explanation:
1. For \( 6x^2\lt1 \): \( \dfrac{5x}{1-6x^2}=1 \), i.e. \( 6x^2+5x-1=0 \), so \( (6x-1)(x+1)=0 \).
2. With \( x\ge0 \) and \( x^2\lt\dfrac16 \), only \( x=\dfrac16 \) works, so \( A \) is a singleton. Answer: (b)

Teacher's Note:
1. Combine the two inverse tangents.

Answer: (a) \( \dfrac9{1+9x^3} \)

Explanation:
1. \( \dfrac{6x^{3/2}}{1-9x^3}=\dfrac{2(3x^{3/2})}{1-(3x^{3/2})^2} \), so the function is \( 2\tan^{-1}(3x^{3/2}) \).
2. Derivative \( =2\cdot\dfrac1{1+9x^3}\cdot\dfrac92x^{1/2}=\sqrt x\cdot\dfrac9{1+9x^3} \). Answer: (a)

Teacher's Note:
1. Recognise \( \tan^{-1}\dfrac{2t}{1-t^2} \) with \( t=3x^{3/2} \).

Answer: (a) \( \dfrac{3x-x^3}{1-3x^2} \)

Explanation:
1. Put \( x=\tan\theta \) with \( |\theta|\lt\dfrac\pi6 \). Then \( \tan^{-1}\dfrac{2x}{1-x^2}=2\theta \), so \( \tan^{-1}y=3\theta \).
2. \( y=\tan3\theta=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=\dfrac{3x-x^3}{1-3x^2} \). Answer: (a)

Teacher's Note:
1. Substitute \( x=\tan\theta \).

Answer: (b) \( \dfrac6{17} \)

Explanation:
1. \( \mathrm{cosec}^{-1}\dfrac53=\tan^{-1}\dfrac34 \) (3-4-5 triangle).
2. \( \tan^{-1}\dfrac34+\tan^{-1}\dfrac23=\tan^{-1}\dfrac{17/12}{1/2}=\tan^{-1}\dfrac{17}6 \), so the cotangent is \( \dfrac6{17} \). Answer: (b)

Teacher's Note:
1. Convert to a single inverse tangent.

Answer: (b) 3

Explanation:
1. \( \mathrm{cosec}^{-1}\dfrac54=\sin^{-1}\dfrac45 \), so \( \sin^{-1}\dfrac x5=\dfrac\pi2-\sin^{-1}\dfrac45=\cos^{-1}\dfrac45 \).
2. \( \cos^{-1}\dfrac45=\sin^{-1}\dfrac35 \), hence \( x=3 \). Answer: (b)

Teacher's Note:
1. Use \( \sin^{-1}t+\cos^{-1}t=\pi/2 \).

Answer: (b) \( 4\sin^2\alpha \)

Explanation:
1. Taking cosine: \( \dfrac{xy}2+\sqrt{1-x^2}\sqrt{1-\frac{y^2}4}=\cos\alpha \), so \( \sqrt{(1-x^2)(4-y^2)}=2\cos\alpha-xy \).
2. Squaring: \( 4-4x^2-y^2+x^2y^2=4\cos^2\alpha-4xy\cos\alpha+x^2y^2 \), so \( 4x^2-4xy\cos\alpha+y^2=4\sin^2\alpha \). Answer: (b)

Teacher's Note:
1. Take cosine of both sides and square.

Answer: (c) \( |a|\le\dfrac1{\sqrt2} \)

Explanation:
1. \( \sin^{-1}x\in\left[-\dfrac\pi2,\dfrac\pi2\right] \), so \( -\dfrac\pi2\le2\sin^{-1}a\le\dfrac\pi2 \).
2. Thus \( -\dfrac\pi4\le\sin^{-1}a\le\dfrac\pi4 \), i.e. \( |a|\le\dfrac1{\sqrt2} \). Answer: (c)

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

Answer: (a) \( \tan^2\left(\dfrac\alpha2\right) \)

Explanation:
1. Adding \( \cot^{-1}t+\tan^{-1}t=\dfrac\pi2 \) gives \( 2\cot^{-1}\sqrt{\cos\alpha}=\dfrac\pi2+x \), so \( \sqrt{\cos\alpha}=\cot\left(\dfrac\pi4+\dfrac x2\right)=\dfrac{\cos\frac x2-\sin\frac x2}{\cos\frac x2+\sin\frac x2} \).
2. Squaring: \( \cos\alpha=\dfrac{1-\sin x}{1+\sin x} \), so by componendo and dividendo \( \sin x=\tan^2\dfrac\alpha2 \). Answer: (a)

Teacher's Note:
1. Add the identity \( \cot^{-1}t+\tan^{-1}t=\pi/2 \).

Answer: (d) \( \tan^{-1}\left(\dfrac12\right) \)

Explanation:
1. \( \tan^{-1}\dfrac14+\tan^{-1}\dfrac29=\tan^{-1}\dfrac{1/4+2/9}{1-1/18}=\tan^{-1}\dfrac{17/36}{17/18}=\tan^{-1}\dfrac12 \).
2. The other options give \( \tan^{-1}\dfrac13 \), \( \tan^{-1}\dfrac13 \) (for (b)) and \( \tfrac12\tan^{-1}\tfrac35\approx0.27 \) versus \( \tan^{-1}\tfrac12\approx0.46 \), so they differ. Answer: (d)

Teacher's Note:
1. Apply the tangent addition formula.