# Question #e50ba

Jul 25, 2016

Given $\tan x = \frac{9}{40} \mathmr{and} x \to \text{In 3rd quadrant}$
So $x = \pi + \alpha . \text{where alpha acute angle}$

$\therefore \tan x = \frac{9}{40} \implies \tan \left(\pi + \alpha\right) = \frac{9}{40}$
$\implies \tan \alpha = \frac{9}{40}$

$\textcolor{red}{2 x = 2 \pi + 2 \alpha \to \text{in 1st or in 2nd quadrant.}}$

$\textcolor{b l u e}{\text{So "sin2x" will be positive.}}$

Now
$\sin 2 x = \sin \left(2 \pi + 2 \alpha\right) = \sin 2 \alpha = \frac{2 \tan \alpha}{1 + {\tan}^{2} \alpha}$

$= \frac{2 \cdot \frac{9}{40}}{1 + {\left(\frac{9}{40}\right)}^{2}}$

$= \frac{2 \cdot \frac{9}{40} \cdot {40}^{2}}{{40}^{2} + {9}^{2}}$

$= \frac{720}{1681}$

Now $\frac{x}{2} = \frac{1}{2} \left(\pi + \alpha\right) = \left(\frac{\pi}{2} + \frac{\alpha}{2}\right)$
$\therefore \tan \left(\frac{x}{2}\right) = \tan \left(\frac{\pi}{2} + \frac{\alpha}{2}\right) = - \cot \left(\frac{\alpha}{2}\right)$

Again $\tan \alpha = \frac{9}{40}$
$\implies \frac{2 \tan \left(\frac{\alpha}{2}\right)}{1 - {\tan}^{2} \left(\frac{\alpha}{2}\right)} = \frac{9}{40}$

If $\tan \left(\frac{\alpha}{2}\right) = a$
then the above relation becomes
$\frac{2 a}{1 - {a}^{2}} = \frac{9}{40}$

$\implies 9 {a}^{2} + 80 a - 9 = 0$

$\implies a = \frac{- 80 + \sqrt{{80}^{2} - 4 \cdot 9 \cdot \left(- 9\right)}}{2 \cdot 9}$

$\textcolor{red}{\left[a > 0 \text{ since " alpha/2 " acute}\right]}$

$= \frac{- 80 + 82}{18} = \frac{2}{18} = \frac{1}{9}$
$\therefore \tan \left(\frac{\alpha}{2}\right) = \frac{1}{9}$
$\implies \cot \left(\frac{\alpha}{2}\right) = 9$

Hence $\tan \left(\frac{x}{2}\right) = - \cot \left(\frac{\alpha}{2}\right) = - 9$