MHT CET 2026 13th April Evening Shift | Inverse Trigonometric Functions Question 25 | Mathematics

The value of $\dfrac{\tan^{-1}(1) + \cos^{-1}\left(\dfrac{1}{2}\right) + \sin^{-1}\left(\dfrac{1}{2}\right)}{\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)}$ is equal to

$\dfrac{3}{4}$

$\dfrac{-1}{3}$

$-4$

$\dfrac{-9}{4}$

MHT CET 2026 13th April Morning Shift

MCQ (Single Correct Answer)

-0

The value of $\tan^{-1}(\sqrt{3}) + \sec^{-1}(-2) - \sin^{-1}\left(-\dfrac{1}{2}\right)$ is

$\dfrac{5\pi}{6}$

$\dfrac{2\pi}{3}$

$\dfrac{7\pi}{6}$

$\dfrac{4\pi}{3}$

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

-0

If $\cot(\cos^{-1} x) = \sec\left(\tan^{-1} \dfrac{a}{\sqrt{b^2 - a^2}}\right)$, then the value of $x$ is

$\dfrac{b}{\sqrt{2b^2 + a^2}}$

$\dfrac{\sqrt{2b^2 - a^2}}{b}$

$\dfrac{\sqrt{2b^2 + a^2}}{b}$

$\dfrac{b}{\sqrt{2b^2 - a^2}}$

MHT CET 2026 11th April Morning Shift

MCQ (Single Correct Answer)

-0

The minimum value of $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ is............

$\dfrac{\pi^2}{8}$

$\dfrac{3\pi^2}{8}$

$\dfrac{5\pi^2}{8}$

$\dfrac{7\pi^2}{8}$

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