Find the value of $\cos \left(\sec ^{-1} x+\csc ^{-1} x\right)$ for $|x| \geq 1$.

If $\sin ^{-1}(4 x)+\sin ^{-1}(4 \sqrt{3} x)=-\frac{\pi}{2}$,then the absolute value of $x$ is

The value of $\frac{\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)}{\operatorname{cosec}^{-1}(-\sqrt{2})+\cos ^{-1}\left(\frac{-1}{2}\right)}$ is

If $\sec ^{-1}\left(\frac{5}{x}\right)+\sin ^{-1} \left(\frac{4}{5}\right)=\frac{\pi}{2}$,where $x \neq 0$,then $x=$ . . . . . . .

If $y = \cos^{-1}(\cos x)$,then find $\frac{dy}{dx}$ at $x = \frac{5\pi}{4}$.

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: