1
MHT CET 2026 16th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\sin^{-1}\left(\dfrac{12}{13}\right) + \cos^{-1}\left(\dfrac{4}{5}\right) + \tan^{-1}\left(\dfrac{63}{16}\right) =$
A
$\dfrac{\pi}{2}$
B
$\dfrac{3\pi}{2}$
C
$\pi$
D
$2\pi$
2
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
The value of $\tan^{-1}\left(\dfrac{\cos\left(\frac{19\pi}{4}\right) - 1}{\sin\left(\frac{\pi}{4}\right)}\right)$ is equal to
A
$-\dfrac{\pi}{4}$
B
$-\dfrac{5\pi}{12}$
C
$-\dfrac{3\pi}{8}$
D
$-\dfrac{4\pi}{9}$
3
MHT CET 2026 15th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $\tan^{-1}\left[\dfrac{\sqrt{5-2\sqrt{6}}}{1+\sqrt{6}}\right] = \dfrac{\pi}{3} - \tan^{-1}(k)$, then $\sec^{-1}(k) = ...$
A
$\dfrac{\pi}{6}$
B
$\dfrac{\pi}{4}$
C
$\dfrac{\pi}{3}$
D
$\dfrac{\pi}{2}$
4
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $x=\tan ^{-1}\left\{\frac{\sqrt{1+t^2}-1}{t}\right\}, y=\cos ^{-1}\left\{\frac{1-t^2}{1+t^2}\right\}, \quad$ then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is equal to

A

2

B

$\frac{1}{2}$

C

4

D

$\frac{1}{4}$

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