Inverse Trigonometric Functions · Mathematics · VITEEE

$S=\cot ^{-1}\left(\frac{1+2 \times 6}{4}\right)+\cot ^{-1}\left(\frac{1+3 \times 8}{10}\right)+\cot ^{-1} \left(\frac{1+4 \times 10}{18}\right) \ldots \ldots \ldots$ upto $\infty$ terms is equal to

The value of $$\sin ^{-1}\left\{\cot \left(\sin ^{-1} \sqrt{\frac{2-\sqrt{3}}{4}}+\cos ^{-1} \frac{\sqrt{12}}{4}+\sec ^{-1} \sqrt{2}\right)\right\}$$ is

The sum of the infinite terms of the series $\cot ^{-1}\left(1^2+\frac{3}{4}\right)+\cot ^{-1}\left(2^2+\frac{3}{4}\right)$ $+\cot ^{-1}\left(3^2+\frac{3}{4}\right)+\ldots$ is equal to

If $$\left(\sin ^{-1} x\right)^2-\left(\cos ^{-1} x\right)^2=a \pi^2$$, then the range of $$a$$ is

The equation $$3 \cos ^{-1} x-\pi x-\pi / 2=0$$ has

If $$\cot ^{-1}(y)=\cot ^{-1}(x)+\cot ^{-1}\left(\frac{x^2-1}{2 x}\right)$$, then the value of $$y$$ is

$$\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$$ equals to

If $$\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$$, where $$|x|< \frac{1}{\sqrt{3}}$$ then value of $$y$$ is

If $$\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$$ where $$-1 \leq x \leq 1, -2 \leq y \leq 2, x \leq \frac{y}{2}$$, then for all $$x, y, 4 x^2-4 x y \cos \alpha+y^2$$ is equal to

Using the principal values, the value of $$\sin ^{-1}\left\{\sin \frac{5 \pi}{6}\right\}+\tan ^{-1}\left\{\tan \frac{\pi}{6}\right\}$$ is equal to

Find the value of $$\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{3}{5}\right)$$.