Question

Solve the equation $(1-\tan x)(1+\sin 2x)=1+\tan x$. x = $n\pi -\frac{\pi }{k},n\pi$, $n\in Z$ then k =

Answer 1

given, $(1-\tan x)(1+\sin 2x)=1+\tan x$

$\Rightarrow (1-\tan x)(1+\frac{2\tan x}{1+{\tan }^{2}x})=1+\tan x$ $[\because \sin 2\theta =\frac{2\tan x}{1+{\tan }^{2}x}]$

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

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

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

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

$\therefore \tan x=-1$ or ${\tan }^{2}x=0$

$x=n\pi -\frac{\pi }{4}$ or $x=n\pi$

Hence solved.