MHT CET 2026 11th April Morning Shift | Inverse Trigonometric Functions Question 29 | Mathematics

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...

$\dfrac{46}{41}$

$\dfrac{23}{41}$

$\dfrac{42}{41}$

$\dfrac{44}{41}$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-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

$\frac{1}{2}$

$\frac{1}{4}$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-0

$$ \cot ^{-1}\left(2 \cdot 1^2\right)+\cot ^{-1}\left(2 \cdot 2^2\right)+\cot ^{-1}\left(2 \cdot 3^2\right)+\ldots \ldots \ldots \infty= $$

$\frac{\pi}{2}$

$\frac{\pi}{3}$

$\frac{\pi}{4}$

$\frac{\pi}{8}$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-0

The value of $\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]$ is

$\frac{5}{4}$

$\frac{5}{16}$

$\frac{-7}{17}$

$\frac{7}{17}$

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