$\tan ^{-1} \sqrt{3}=\tan ^{-1} \tan \left(\frac{\pi}{3}\right)$

$\left[\because \tan ^{-1}(\tan x)=x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right]=\left(\frac{\pi}{3}\right)$

We have, $$\cos ^{-1}\left(\cos \frac{14 \pi}{3}\right)=\cos ^{-1} \cos \left(4 \pi+\frac{2 \pi}{3}\right)$$

$$\begin{array}{lr} =\cos ^{-1} \cos \frac{2 \pi}{3} & {[\because \cos (2 n \pi+\theta)=\cos \theta]} \\ =\frac{2 \pi}{3} & \left\{\because \cos ^{-1}(\cos x)=x, x \in[0, \pi]\right\} \end{array}$$

$$\begin{aligned} &\begin{gathered} \cos \left(\sin ^{-1} x+\cos ^{-1} x\right) \\ =\cos \frac{\pi}{2}=0 \end{gathered}\\ &\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right] \end{aligned}$$

$\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$

$\tan \left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right)=\tan \left(\frac{\pi / 2}{2}\right)$

$=\tan \frac{\pi}{4}=1$

$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} & y=2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^2} \\ \therefore\quad & y=2 \tan ^{-1} \tan \theta+\sin ^{-1} \frac{2 \tan \theta}{1+\tan ^2 \theta} \quad \quad[\text { let } x=\tan \theta] \end{aligned} \end{aligned}$$

$\Rightarrow \quad y=2 \theta+\sin ^{-1} \sin 2 \theta\quad \left[\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta}\right]$

$\Rightarrow \quad y=2 \theta+2 \theta=4 \theta \quad\left[\because \theta=\tan ^{-1} x\right]$

$$\begin{array}{ll} \Rightarrow & y=4 \tan ^{-1} x \\ \because & -\pi / 2<\tan ^{-1} x<\pi / 2 \\ \therefore & -\frac{4 \pi}{2}<4 \tan ^{-1} x<4 \pi / 2 \\ \Rightarrow & -2 \pi<4 \tan ^{-1} x<2 \pi \\ \Rightarrow & -2 \pi< y< 2 \pi \quad \left[\because y=4 \tan ^{-1} x\right] \end{array}$$