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1
MHT CET 2026 11th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
If $\cot(\cos^{-1} x) = \sec\left(\tan^{-1} \dfrac{a}{\sqrt{b^2 - a^2}}\right)$, then the value of $x$ is
A
$\dfrac{b}{\sqrt{2b^2 + a^2}}$
B
$\dfrac{\sqrt{2b^2 - a^2}}{b}$
C
$\dfrac{\sqrt{2b^2 + a^2}}{b}$
D
$\dfrac{b}{\sqrt{2b^2 - a^2}}$
2
MHT CET 2026 11th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
The minimum value of $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ is............
A
$\dfrac{\pi^2}{8}$
B
$\dfrac{3\pi^2}{8}$
C
$\dfrac{5\pi^2}{8}$
D
$\dfrac{7\pi^2}{8}$
3
MHT CET 2026 11th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $\tan^{-1}(1) + \tan^{-1}(3) + \tan^{-1}(5) + \tan^{-1}\left(\dfrac{1}{4}\right) = \pi + \tan^{-1}\left(\dfrac{\alpha}{2}\right)$, then the value of $\alpha$ is...
A
$\dfrac{46}{41}$
B
$\dfrac{23}{41}$
C
$\dfrac{42}{41}$
D
$\dfrac{44}{41}$
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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