KCET 2026 | Inverse Trigonometric Functions Question 1 | Mathematics

$\tan^{-1}\left(\dfrac{1}{1 + 1 \cdot 2}\right) + \tan^{-1}\left(\dfrac{1}{1 + 2 \cdot 3}\right) + \ldots + \tan^{-1}\left(\dfrac{1}{1 + n(n+1)}\right) = $

$\tan^{-1}\left(\dfrac{n}{n+2}\right)$

$\tan^{-1}\left(\dfrac{n+1}{n}\right)$

$\tan^{-1}\left(\dfrac{n}{n+1}\right)$

$\tan^{-1}\left(\dfrac{n+2}{n}\right)$

KCET 2026

MCQ (Single Correct Answer)

-0

If $\sin^{-1} x + \sin^{-1} y = \dfrac{\pi}{2}$, then $x^2$ is equal to

$1 - y^2$

$\sqrt{1 - y^2}$

$-\sqrt{1 - y^2}$

$1 + y^2$

KCET 2025

MCQ (Single Correct Answer)

-0

$$ \sec ^2\left(\tan ^{-1} 2\right)+\operatorname{cosec}^2\left(\cot ^{-1} 3\right)= $$

KCET 2025

MCQ (Single Correct Answer)

-0

$2 \cos ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$ is valid for all values of ' $x$ ' satisfying

$0 \leq x \leq \frac{1}{\sqrt{2}}$

$-1 \leq x \leq 1$

$0 \leq x \leq 1$

$\frac{1}{\sqrt{2}} \leq x \leq 1$

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