# Solve for a: 2 tan x - tan 2x +2a=1-tan2x tan^2 x ?

Nov 22, 2017

$a = \frac{1}{2.}$

#### Explanation:

We know that, $\tan 2 x = \frac{2 \tan x}{1 - {\tan}^{2} x} = \frac{2 t}{1 - {t}^{2}} , w h e r e , t = \tan x .$

$\therefore 2 \tan x - \tan 2 x + 2 a = 1 - \tan 2 x {\tan}^{2} x ,$

$\Rightarrow 2 t - \frac{2 t}{1 - {t}^{2}} + 2 a = 1 - \frac{2 t}{1 - {t}^{2}} \cdot {t}^{2.}$

Multiplying by $\left(1 - {t}^{2}\right) ,$ we get,

$2 t \left(1 - {t}^{2}\right) - 2 t + 2 a \left(1 - {t}^{2}\right) = \left(1 - {t}^{2}\right) - 2 {t}^{3.}$

$\therefore \cancel{2 t - 2 {t}^{3} - 2 t} + 2 a \left(1 - {t}^{2}\right) = 1 - {t}^{2} \cancel{- 2 {t}^{3}} , \mathmr{and} ,$

$2 a \left(1 - {t}^{2}\right) = \left(1 - {t}^{2}\right) .$

$\therefore 2 a = \frac{1 - {t}^{2}}{1 - {t}^{2}} = 1 , \mathmr{if} {t}^{2} \ne 1.$

$\therefore a = \frac{1}{2} , \mathmr{if} {t}^{2} \ne 1.$

Now, ${t}^{2} = 1 \Rightarrow t = \tan x = \pm 1 = \tan \left(\pm \frac{\pi}{4}\right) ,$

$\Rightarrow x = n \pi \pm \frac{\pi}{4} , n \in \mathbb{Z} ,$ but, then, $\tan 2 x$ becomes undefined.

$\therefore t = \tan x \ne \pm 1.$

$\therefore a = \frac{1}{2.}$