1
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $\alpha>\beta>\gamma>0$, then the expression $\cot ^{-1}\left\{\beta+\frac{\left(1+\beta^2\right)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\frac{\left(1+\gamma^2\right)}{(\beta-\gamma)}\right\}+\cot ^{-1}\left\{\alpha+\frac{\left(1+\alpha^2\right)}{(\gamma-\alpha)}\right\}$ is equal to :

A
$3 \pi$
B
$\frac{\pi}{2}-(\alpha+\beta+\gamma)$
C
$\pi$
D
0
2
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}$, then $\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)$ is equal to

A
$x+\tan ^{-1} \frac{5}{12}$
B
$x-\tan ^{-1} \frac{4}{3}$
C
$x+\tan ^{-1} \frac{4}{5}$
D
$x-\tan ^{-1} \frac{5}{12}$
3
JEE Main 2025 (Online) 22nd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right)$ is :

A
$24 \pi^2$
B
$18 \pi^2$
C
$22 \pi^2$
D
$31 \pi^2$
4
JEE Main 2024 (Online) 4th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Given that the inverse trigonometric function assumes principal values only. Let $$x, y$$ be any two real numbers in $$[-1,1]$$ such that $$\cos ^{-1} x-\sin ^{-1} y=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi$$. Then, the minimum value of $$x^2+y^2+2 x y \sin \alpha$$ is

A
0
B
$$-$$1
C
$$\frac{1}{2}$$
D
$$\frac{-1}{2}$$
JEE Main Subjects
EXAM MAP