If sin x=

Given:

$\sin x=\frac{\sqrt{5}}{3}$

Using the identity $\cos x=\sqrt{1-\sin ^{2} x}$, we get

$\cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\left(\frac{\sqrt{5}}{3}\right)^{2}}=\pm \frac{2}{3}$

Since x lies in the 2nd quadrant, cosx is negative.

$\therefore \cos x=-\frac{2}{3}$

Now,

$\cos x=1-2 \sin ^{2} \frac{x}{2}$

$\Rightarrow-\frac{2}{3}=1-2 \sin ^{2} \frac{x}{2}$

$\Rightarrow \sin \frac{x}{2}=\pm \sqrt{\frac{5}{6}}$

Since $x$ lies in the 2 nd quadrant, $\frac{x}{2}$ lies in the 1 st quadrant.

$\therefore \sin \frac{x}{2}=\sqrt{\frac{5}{6}}$

Again,

$\cos x=2 \cos ^{2} \frac{x}{2}-1$

$\Rightarrow-\frac{2}{3}=2 \cos ^{2} \frac{x}{2}-1$

$\Rightarrow \cos \frac{x}{2}=\pm \frac{1}{\sqrt{6}}$

$\Rightarrow \cos \frac{x}{2}=\frac{1}{\sqrt{6}} \quad\left(\because \frac{x}{2}<\frac{\pi}{2}\right)$

Now, $\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\frac{\sqrt{5}}{\sqrt{6}}}{\frac{1}{\sqrt{6}}}=\sqrt{5}$