$\text { Let } y=\cot ^{-1}\left(\frac{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}+\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}-\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}\right]$
$1+\sin \left(\frac{4 x}{3}\right)=1+\cos \left(\frac{\pi}{2}-\frac{4 x}{3}\right)=2 \cos ^2\left(\frac{\pi}{4}-\frac{2 x}{3}\right)$
$\therefore \sqrt{1+\sin \left(\frac{4 x}{3}\right)}=\sqrt{2} \cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)$
Also, $1-\sin \left(\frac{4 x}{3}\right)=1-\cos \left(\frac{\pi}{2}-\frac{4 x}{3}\right)=2 \sin ^2\left(\frac{\pi}{4}-\frac{2 x}{3}\right)$
$\therefore \sqrt{1-\sin \left(\frac{4 x}{3}\right)}=\sqrt{2} \sin \left(\frac{\pi}{4}-\frac{2 x}{3}\right)$
$\therefore \frac{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}+\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}-\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}$
$=\frac{\sqrt{2} \cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)+\sqrt{2} \sin \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}{\sqrt{2} \cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)-\sqrt{2} \sin \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}$
$=\frac{\cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)+\sin \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}{\cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)-\sin \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}$
$=\frac{1+\tan \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}{1-\tan \left(\frac{\pi}{4}-\frac{2 x}{3}\right)} \quad \cdots\left[\right.$ Dividing by $\left.\cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)\right]$
$=\frac{\tan \frac{\pi}{4}+\tan \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}{1-\tan \frac{\pi}{4} \cdot \tan \left(\frac{\pi}{4}-\frac{2 x}{3}\right)} \quad \cdots\left[\because \tan \frac{\pi}{4}=1\right]$
$=\tan \left[\frac{\pi}{4}+\frac{\pi}{4}-\frac{2 x}{3}\right]=\tan \left(\frac{\pi}{2}-\frac{2 x}{3}\right)$
$=\cot \left(\frac{2 x}{3}\right)$
$\therefore y=\cot ^{-1}\left[\cot \left(\frac{2 x}{3}\right)\right]=\frac{2 x}{3}$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x} =\frac{d}{d x}\left(\frac{2 x}{3}\right)=\frac{2}{3} \frac{d}{d x}(x)$
$=\frac{2}{3} \times 1=\frac{2}{3} .$

Maths Solve the Following Question.(4 Marks)

Solve the Following Question.(4 Marks)