The value of 2^2(^(-1)2)+3cosec^2(^(-1)3) is: Option 1: 15 Option 2: 30 Option 3: 40 Option 4: 10

Answer: 40. Solution: ^2(^(-1)2)=1+^2(^(-1)2)=1+4=5 and cosec^2(^(-1)3)=1+3^2=10. So the value is 2(5)+3(10)=40.

Geometry

Trigonometry

Easy

core

The value of $2\sec^2(\tan^{-1}2)+3\text{cosec}^2(\cot^{-1}3)$ is:

✅ Correct Option: 3

$\sec^2(\tan^{-1}2)=1+\tan^2(\tan^{-1}2)=1+4=5$ and $\text{cosec}^2(\cot^{-1}3)=1+3^2=10$ . So the value is $2(5)+3(10)=40$ .

2022: 10 Aug Shift 1

The value of $\frac{1}{AB^2} + \frac{1}{BC^2}$ is:

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 :

2023: 15 June Shift 2

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

2025: 15 May Shift 1

For the principal value branch, the value of $\sin\left(\frac{\pi}{2} - \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)$ is

2022: 10 Aug Shift 1

$\angle ABC$ is equal to:

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: