Match the LIST-I with LIST-II, LIST-I, LIST-II, A., If ^(-1) + ^(-1) + ^(-1) = 3 then (2+) + (2+) + (2+) equals, I., 1, B., The value of ^2(^(-1)4) + ^2(cosec^(-1)5) is, II., 9, C., If (^(-1)2/5 + ^(-1)x) = 1 then 10x equals, III., 4, D., The length of the interval of domain of ^(-1)2x is, IV., 39, Choose the correct answer from the options given below: Option 1: A - II, B - I, C - III, D - IV Option 2: A - II, B - III, C - IV, D - I Option 3: A - II, B - IV, C - III, D - I Option 4: A - I, B - IV, C - III, D - II

Answer: A - II, B - IV, C - III, D - I. Solution: A: the sum is 3 only when each inverse cosine is, so ===-1 and the expression equals 9 (II). B: ^2(^(-1)4)=15 and ^2(^(-1)5)=24, total 39 (IV). C: the argument must be 2, so ^(-1)x=^(-1)25, giving 10x=4 (III). D: -12≤ x≤12 has length 1 (I).

Geometry

Trigonometry

Hard

core

Match the LIST-I with LIST-II

LIST-I LIST-II
A. If $\cos^{-1}\alpha + \cos^{-1}\beta + \cos^{-1}\gamma = 3\pi$ then $\alpha(2\beta+\gamma) + \beta(2\gamma+\alpha) + \gamma(2\alpha+\beta)$ equals I. $1$
B. The value of $\tan^2(\sec^{-1}4) + \cot^2(\text{cosec}^{-1}5)$ is II. $9$
C. If $\sin\left(\cos^{-1}\dfrac{2}{5} + \sin^{-1}x\right) = 1$ then $10x$ equals III. $4$
D. The length of the interval of domain of $\sin^{-1}2x$ is IV. $39$

Choose the correct answer from the options given below:

	LIST-I		LIST-II
A.	If $\cos^{-1}\alpha + \cos^{-1}\beta + \cos^{-1}\gamma = 3\pi$ then $\alpha(2\beta+\gamma) + \beta(2\gamma+\alpha) + \gamma(2\alpha+\beta)$ equals	I.	$1$
B.	The value of $\tan^2(\sec^{-1}4) + \cot^2(\text{cosec}^{-1}5)$ is	II.	$9$
C.	If $\sin\left(\cos^{-1}\dfrac{2}{5} + \sin^{-1}x\right) = 1$ then $10x$ equals	III.	$4$
D.	The length of the interval of domain of $\sin^{-1}2x$ is	IV.	$39$

A - II, B - I, C - III, D - IV

A - II, B - III, C - IV, D - I

A - II, B - IV, C - III, D - I

A - I, B - IV, C - III, D - II

✅ Correct Option: 3

A: the sum is $3\pi$ only when each inverse cosine is $\pi$ , so $\alpha=\beta=\gamma=-1$ and the expression equals 9 (II). B: $\tan^2(\sec^{-1}4)=15$ and $\cot^2(\csc^{-1}5)=24$ , total 39 (IV). C: the argument must be $\frac\pi2$ , so $\sin^{-1}x=\sin^{-1}\frac25$ , giving $10x=4$ (III). D: $-\frac12\le x\le\frac12$ has length 1 (I).

2026: 21 May Shift 2

Match the LIST-I with LIST-II

LIST-I (Function)		LIST-II (Range of the principal value branch)
A. $\cos^{-1}x$		I. $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
B. $\text{cosec}^{-1}x$		II. $(0, \pi)$
C. $\tan^{-1}x$		III. $[0, \pi]$
D. $\cot^{-1}x$		IV. $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}$

Choose the correct answer from the options given below:

2022: 25 May Shift 1

Match List I with List II

LIST I	LIST II
A. $\sin^{-1} x + \cos^{-1} x, x \in [-1,1]$	I. $-\frac{\pi}{2}$
B. $\tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3})$	II. $-\frac{\pi}{6}$
C. $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)$	III. $\frac{\pi}{2}$
D. $\sin^{-1}\left(-\frac{1}{2}\right)$	IV. $\frac{\pi}{6}$

Choose the correct answer from the options given below:

2022: 30 Aug Shift 1

The value of $\sin\left[2\cot^{-1}\left(\frac{-5}{12}\right)\right]$ is :

2022: 10 Aug Shift 1

$\angle ABC$ is equal to:

2025: 22 May Shift 2

The value of $\cot\left(\cos^{-1}\frac{7}{25}\right)$ is

2022: 4 Aug Shift 1

Match List - I with List - II

List - I	List - II
A. $\tan^{-1}\left(\tan\frac{7\pi}{6}\right)$	I. $\frac{5\pi}{6}$
B. $\tan^{-1}\left(\tan\frac{8\pi}{6}\right)$	II. $\frac{\pi}{2}$
C. $\tan^{-1}\frac{1}{\sqrt{3}} + cosec^{-1}\frac{2}{\sqrt{3}}$	III. $\frac{\pi}{6}$
D. $\cos^{-1}\left(\cos\frac{5\pi}{6}\right)$	IV. $\frac{\pi}{3}$

Choose the correct answer from the options given below:

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