If $(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$,then the value of $x$ is

Find $\frac{dy}{dx}$ for $y = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ where $|x| < 1$.

The real value of $x$ that satisfies the equation $\tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4}$ is

$\cosh \left(\sinh ^{-1}(\sqrt{8})+\cosh ^{-1} 5\right)=$

Consider the following statements:
Assertion $(A)$: For $x \in \mathbb{R}-\{1\}$, $\frac{d}{dx}\left(\tan^{-1}\left(\frac{1+x}{1-x}\right)\right) = \frac{d}{dx}\left(\tan^{-1} x\right)$.
Reason $(R)$: For $x < 1$, $\tan^{-1}\left(\frac{1+x}{1-x}\right) = \frac{\pi}{4} + \tan^{-1} x$, and for $x > 1$, $\tan^{-1}\left(\frac{1+x}{1-x}\right) = -\frac{3\pi}{4} + \tan^{-1} x$.
The correct answer is:

If $\theta = \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18)$,then $\cot \theta$ is equal to