Question

if $\tan x=\frac{3}{4}$ and $x$ lies in the $III$ quadrant, find the value of $\sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$

Answer 1

Given that $\tan x=\frac{3}{4}$

We know that $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$

Applying this formula we get

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

$\Rightarrow \frac{3}{4}=\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}$

$\Rightarrow 3-3{\tan }^2\frac{x}{2}=8\tan \frac{x}{2}$

$\Rightarrow 3{\tan }^2\frac{x}{2}+8\tan \frac{x}{2}-3=0$

$\Rightarrow 3{\tan }^2\frac{x}{2}+9\tan \frac{x}{2}-\tan \frac{x}{2}-3=0$

$\Rightarrow 3\tan \frac{x}{2}(\tan \frac{x}{2}+3)-(\tan \frac{x}{2}+3)=0$

$\Rightarrow (3\tan \frac{x}{2}-1)(\tan \frac{x}{2}+3)=0$

$\Rightarrow \tan \frac{x}{2}=-3,\tan \frac{x}{2}=\frac{1}{3}$

Given that $x$ lies in $Q_3$

$\Rightarrow {180}^0\le x\le {270}^0$

$\Rightarrow {90}^0\le \frac{x}{2}\le {135}^0$

Thus $\frac{x}{2}$ lies in $Q_2$

Therefore, $\tan \frac{x}{2}$ is negative

Thus, $\tan \frac{x}{2}=-3$

Therefore $\sin \frac{x}{2}=\frac{3}{\sqrt{10}}$ and $\cos \frac{x}{2}=\frac{-1}{\sqrt{10}}$