^(-1)( x) = ()/2 - x is valid for : Option 1: ≤ x ≤ 0 Option 2: 0 ≤ x ≤ Option 3: -()/2 ≤ x ≤ ()/2 Option 4: -()/4 ≤ x ≤ (3)/4

Answer: 0 ≤ x ≤. Solution: ^(-1)( x) = ^(-1)((()/2 - x)) = ()/2 - x when ()/2 - x [-()/2, ()/2], i.e., 0 ≤ x ≤ .

Geometry

Trigonometry

Medium

$\sin^{-1}(\cos x) = \frac{\pi}{2} - x$ is valid for :

$-\pi \leq x \leq 0$

$0 \leq x \leq \pi$

$-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$

$-\frac{\pi}{4} \leq x \leq \frac{3\pi}{4}$

✅ Correct Option: 2

$\sin^{-1}(\cos x) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - x\right)\right) = \frac{\pi}{2} - x$ when $\frac{\pi}{2} - x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , i.e., $0 \leq x \leq \pi$ .

2022: 10 Aug Shift 1

$\angle ABC$ is equal to:

2025: 3 June Shift 2

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

2026: 18 May Shift 2

Match the LIST-I with LIST-II

LIST-I Functions	LIST-II Principal value
A. $\sin^{-1}\left(\dfrac{1}{2}\right)$	I. $\dfrac{\pi}{4}$
B. $\sin^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)$	II. $\dfrac{\pi}{6}$
C. $\tan^{-1}(1)$	III. $-\dfrac{\pi}{4}$
D. $\cos^{-1}\left(-\dfrac{1}{2}\right)$	IV. $\dfrac{2\pi}{3}$

Choose the correct answer from the options given below:

2022: 10 Aug Shift 1

$\tan(\alpha + \beta) =$

2026: 22 May Shift 2

$\cos[\tan^{-1}\{\sin(\cot^{-1}x)\}]$ is equal to

2022: 25 May Shift 1

Value of $\frac{e^{\sin(\tan^{-1} x + \cot^{-1} x)}}{e^{\sin(\sin^{-1} x + \cos^{-1} x)}}, x \in [-1, 1]$ , is:

sin⁡−1(cos⁡x)=π2−x\sin^{-1}(\cos x) = \frac{\pi}{2} - xsin−1(cosx)=2π​−x is valid for :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\sin^{-1}(\cos x) = \frac{\pi}{2} - x$ is valid for :