If ${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z=\pi$, then -

Important Questions on Inverse Trigonometric Functions

EASY

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IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-1>Q 62

If ${\tan }^{-1}\frac{x}{\pi }<\frac{\pi }{3}, x\in N$, then the maximum value of $x$ is

MEDIUM

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IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-1>Q 63

The solution of the inequality ${\left({\tan }^{-1}x\right)}^2-3{\tan }^{-1}x+2\ge 0$ is -

HARD

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IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-1>Q 64

The set of values of $x$, satisfying the equation ${\tan }^2\left({\sin }^{-1}x\right)>1$ is -

HARD

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IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-2>Q 1

${\cos }^{-1}x={\tan }^{-1}x$ then -

HARD

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IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-2>Q 2

The value of $\sin \left(\frac{1}{2}{\cot }^{-1}\left(-\frac{3}{4}\right)\right)+\cos \left(\frac{1}{2}{\cot }^{-1}\left(-\frac{3}{4}\right)\right)$ is/are equal to -

MEDIUM

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IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-2>Q 3

The value of ${\tan }^{-1}\left(\frac{1}{2}\tan 2A\right)+{\tan }^{-1}\left(\cot A\right)+{\tan }^{-1}\left({\cot }^{3}A\right)$ for $0<A<\left(\pi /4\right)$ is -

HARD

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IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-2>Q 4

For the equation $2x=\tan \left(2{\tan }^{-1}a\right)+2\tan \left({\tan }^{-1}a+{\tan }^{-1}{a}^3\right)$, which of the following is/are invalid?

HARD

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IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-2>Q 5

The value of ${\left[\tan \left\{\frac{\pi }{4}+\frac{1}{2}{\sin }^{-1}\left(\frac{a}{b}\right)\right\}+\tan \left\{\frac{\pi }{4}-\frac{1}{2}{\sin }^{-1}\left(\frac{a}{b}\right)\right\}\right]}^{-1}$, where $\left(0<a<b\right)$, is -