The value of $\cos (\tan ^{ - 1}(\tan 2))$ is

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:

Evaluate: $\cos^{-1}(\cos \frac{4\pi}{3}) + \sin^{-1}(\sin \frac{4\pi}{3}) = \dots$

If $\tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5}$,then $\cot ^{-1} x+\cot ^{-1} y$ is equal to

If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}$,then $\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)$ is equal to

Prove $\tan ^{-1}\left(\frac{2}{11}\right)+\tan ^{-1}\left(\frac{7}{24}\right)=\tan ^{-1}\left(\frac{1}{2}\right)$