Question #f3a89

Nov 24, 2016

Given $\tan x = \frac{4}{5} \mathmr{and} \pi < x < 2 \pi$

So $\frac{\pi}{2} < \frac{x}{2} < \pi \implies \frac{x}{2} \in \text{ 2nd quadrant}$

Hence $\tan \left(\frac{x}{2}\right) \to - v e$

Now given formula is

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

Substituting $x \text{ for } \frac{x}{2}$ we get

$\tan x = \frac{2 \tan \left(\frac{x}{2}\right)}{1 - {\tan}^{2} \left(\frac{x}{2}\right)}$

$\implies \frac{4}{5} = \frac{2 \tan \left(\frac{x}{2}\right)}{1 - {\tan}^{2} \left(\frac{x}{2}\right)}$

Let $\tan \left(\frac{x}{2}\right) = y$

$\implies {\cancel{4}}^{2} / 5 = \frac{\cancel{2} y}{1 - {y}^{2}}$

$\implies 5 y = 2 - 2 {y}^{2}$

$\implies 2 {y}^{2} + 5 y - 2 = 0$

$\implies y = \frac{- 5 - \sqrt{{5}^{2} - 4 \cdot 2 \cdot \left(- 2\right)}}{2 \cdot 2}$

$\implies y = \frac{- 5 - \sqrt{41}}{4}$

$\implies \tan \left(\frac{x}{2}\right) = \frac{- 5 - \sqrt{41}}{4}$

[since $\tan \left(\frac{x}{2}\right)$ to be negative as explained above +ve root of y is neglected]