The value of [2^(-1)((-5)/12)] is : Option 1: 120/169 Option 2: (-120)/169 Option 3: (-60)/169 Option 4: 60/169

Answer: (-120)/169. Solution: Let = ^(-1)(-5/12), with (/2,). Then = -5/13 and = 12/13. 2 = 2 = 212/13(-5/13) = -120/169.

Geometry

Trigonometry

Medium

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

$\frac{120}{169}$

$\frac{-120}{169}$

$\frac{-60}{169}$

$\frac{60}{169}$

✅ Correct Option: 2

Let $\theta = \cot^{-1}\left(-\frac{5}{12}\right)$ , with $\theta \in (\pi/2,\pi)$ . Then $\cos\theta = -\frac{5}{13}$ and $\sin\theta = \frac{12}{13}$ .

$\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\frac{12}{13}\cdot\left(-\frac{5}{13}\right) = -\frac{120}{169}$ .

2025: 19 May Shift 1

The maximum value of $\sin x \cdot \cos x$ is:

2023: 15 June Shift 2

Let $\tan^{-1}y = \tan^{-1}x + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ . Then $y$ is :

2026: 22 May Shift 1

Match List-I with List-II

List-I (Function)	List-II (Principal value of y)
(A) $y=\sin^{-1}\left(-\frac{1}{2}\right)$	(I) $\frac{3\pi}{4}$
(B) $y=\cot^{-1}\left(-\frac{1}{\sqrt3}\right)$	(II) $-\frac{\pi}{4}$
(C) $y=cosec^{-1}(-\sqrt2)$	(III) $-\frac{\pi}{6}$
(D) $y=\cos^{-1}\left(-\frac{1}{\sqrt2}\right)$	(IV) $\frac{2\pi}{3}$

Choose the correct answer from the options given below:

2026: 30 May Shift 2

Match List-I with List-II

List-I	List-II
(A) Domain of $\sin^{-1}(2x-3)$	(I) [-1, 0]
(B) Domain of $\sin^{-1}(1+2x)$	(II) [-1, 1]
(C) Domain of $\cos^{-1}(2x-1)$	(III) [1, 2]
(D) Domain of $\sin^{-1}(x^2)$	(IV) [0, 1]

Choose the correct answer from the options given below:

2022: 16 July Shift 2

The graph of $\sin^{-1} x$ is represented by:

2023: 22 May Shift 3

The principal value of $\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$ is :

The value of sin⁡[2cot⁡−1(−512)]\sin\left[2\cot^{-1}\left(\frac{-5}{12}\right)\right]sin[2cot−1(12−5​)] is :

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

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