Question

Find the general solutions of $\sin x.\tan x=\tan x-\sin x+1$

Answer 1

$\sin x\tan x-\tan x+\sin x-1=0$

$\tan x(\sin x-1)+1(\sin x-1)=0$

$\Rightarrow (\sin x-1)(\tan x+1)=0$

$\Rightarrow \sin x=1,\tan x=-1$

$\sin x=1\Rightarrow x=2n\pi +\frac{\pi }{2}$

$\tan x=-1\Rightarrow x=n\pi -\frac{\pi }{4}$

But when $x=\frac{\pi }{2}$ and multiples of $\frac{\pi }{2},\tan x$ is not defined.

$\therefore x=n\pi -\frac{\pi }{4}$ is the general solution.