Let ^(-1)y = ^(-1)x + ^(-1)(2x/(1-x^2)). Then y is : Option 1: (3x - x^3)/(1 - 3x^2) Option 2: (3x + x^3)/(1 - 3x^2) Option 3: (3x - x^3)/(1 + 3x^2) Option 4: (3x + x^3)/(1 + 3x^2)

Answer: (3x - x^3)/(1 - 3x^2). Solution: Since ^(-1)(2x/(1-x^2)) = 2^(-1)x: ^(-1)y = 3^(-1)x, so y = (3) where = ^(-1)x. y = (3 - ^3)/(1 - 3^2) = (3x - x^3)/(1 - 3x^2).

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

$\frac{3x - x^3}{1 - 3x^2}$

$\frac{3x + x^3}{1 - 3x^2}$

$\frac{3x - x^3}{1 + 3x^2}$

$\frac{3x + x^3}{1 + 3x^2}$

✅ Correct Option: 1

Since $\tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}x$ :

$\tan^{-1}y = 3\tan^{-1}x$ , so $y = \tan(3\theta)$ where $\theta = \tan^{-1}x$ .

$y = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \frac{3x - x^3}{1 - 3x^2}$ .

2025: 30 May Shift 1

Consider $f(x) = \sin(3x) + 4, \forall x \in \mathbb{R}$ . Then

(A) Maximum value of $f(x)$ is 5

(B) Minimum value of $f(x)$ is 3

(D) Minimum value of $f(x)$ is attained at $x = 0$

Choose the correct answer from the options given below:

2025: 22 May Shift 2

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

2023: 23 May Shift 3

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

2025: 21 May Shift 2

$\sin^{-1}(\cos\frac{3\pi}{5})$ equals

2025: 13 May Shift 1

The value of $\tan^{-1}(1) + \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right)$ is

2022: 10 Aug Shift 1

$\angle ABC$ is equal to:

Let tan⁡−1y=tan⁡−1x+tan⁡−1(2x1−x2)\tan^{-1}y = \tan^{-1}x + \tan^{-1}\left(\frac{2x}{1-x^2}\right)tan−1y=tan−1x+tan−1(1−x22x​). Then yyy is :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

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