If $\tan x=\frac{3}{4}, \pi < x < \frac{3 \pi}{2},$ find the value of $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$

since $\pi < x < \frac{3 \pi}{2}, \cos x$ is negative.

Also $\quad \frac{\pi}{2}<\frac{x}{2}<\frac{3 \pi}{4}$

Therefore, $\sin \frac{x}{2}$ is positive and $\cos \frac{x}{2}$ is negative.

Now $\sec ^{2} x=1+\tan ^{2} x=1+\frac{9}{16}=\frac{25}{16}$

Therefore $\cos ^{2} x=\frac{16}{25}$ or $\cos x=-\frac{4}{5} \quad$ (Why?)

Now $2 \sin ^{2} \frac{x}{2}=1-\cos x=1+\frac{4}{5}=\frac{9}{5}$

Therefore $\sin ^{2} \frac{x}{2}=\frac{9}{10}$

or $\sin \frac{x}{2}=\frac{3}{\sqrt{10}} \quad$ (Why?)

Again $2 \cos ^{2} \frac{x}{2}=1+\cos x=1-\frac{4}{5}=\frac{1}{5}$

Therefore $\cos ^{2} \frac{x}{2}=\frac{1}{10}$or $\cos \frac{x}{2}=-\frac{1}{\sqrt{10}}(\text { Why? })$

Hence $\quad \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{3}{\sqrt{10}} \times\left(\frac{-\sqrt{10}}{1}\right)=-3$