The simplest form of ^(-1) (√(1+x^2) - 1)/x, x ≠ 0 is: Option 1: ^(-1) x Option 2: x Option 3: 1/2 ^(-1) x Option 4: √(1+x^2)

Answer: 1/2 ^(-1) x. Solution: Let x =, so √(1+x^2) = . Then ( - 1)/ = (1 - )/ = (/2). Hence ^(-1)((/2)) = /2 = (1/2)^(-1)x.

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The simplest form of $\tan^{-1} \frac{\sqrt{1+x^2} - 1}{x}$ , $x \neq 0$ is:

$\tan^{-1} x$

$x$

$\frac{1}{2} \tan^{-1} x$

$\sqrt{1+x^2}$

✅ Correct Option: 3

Let $x = \tan\theta$ , so $\sqrt{1+x^2} = \sec\theta$ . Then $(\sec\theta - 1)/\tan\theta = (1 - \cos\theta)/\sin\theta = \tan(\theta/2)$ . Hence $\tan^{-1}(\tan(\theta/2)) = \theta/2 = (1/2)\tan^{-1}x$ .

2022: 17 Aug Shift 2

Match List - I with List - II.

List - I	List - II
(A) $\tan^{-1}\sqrt{3} - \sec^{-1}(-2)$	(I) $\frac{3\pi}{4}$
(B) $\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$	(II) $-\frac{\pi}{3}$
(C) $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$	(III) $\frac{\pi}{2}$
(D) $\cos^{-1}\left(\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right)$	(IV) $\frac{2\pi}{3}$

Choose the correct answer from the options given below :

2025: 3 June Shift 2

The value of $\tan^{-1}(2) + \tan^{-1}(3)$ is equal to

2023: 30 May Shift 3

If $\sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3}$ , then the value of $\cos^{-1} x + \cos^{-1} y$ is :

2025: 27 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $\tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24}$	(I) $\frac{3\pi}{4}$
(B) $\tan^{-1}2 + \tan^{-1}3$	(II) $\pi$
(C) $\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3$	(III) $\tan^{-1}\frac{1}{2}$
(D) $\tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{13}$	(IV) $\tan^{-1}\frac{2}{9}$

Choose the correct answer from the options given below:

2023: 23 May Shift 3

The maximum value of $\sin x + \cos x, x \in R$ is:

2025: 3 June Shift 1

The value of $-\cosec^2(\cot^{-1}y) + \sec^2( \tan^{-1}x)$ is equal to

The simplest form of tan⁡−11+x2−1x\tan^{-1} \frac{\sqrt{1+x^2} - 1}{x}tan−1x1+x2​−1​, x≠0x \neq 0x=0 is:

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CUET Mathematics Past Year Questions (Topic-Wise)

The simplest form of $\tan^{-1} \frac{\sqrt{1+x^2} - 1}{x}$ , $x \neq 0$ is: