# Solve the equation $\sin x\tan x-1=\tan x-\sin x$

$\begin{array}{1 1}(A)\;n\pi+(-1)^n \large\frac{\pi}{2}&(B)\;2n\pi+(-1)^n \large\frac{\pi}{4}\\(C)\;n\pi+(-1)^n \large\frac{\pi}{6}&(D)\;2n\pi+ \large\frac{3\pi}{4}\end{array}$

Toolbox:
• $\tan \theta=\tan\alpha\Rightarrow \theta=n\pi+\alpha$
• $\sin \theta=\sin \alpha\Rightarrow \theta=n\pi+(-1)^n\alpha$
Given
$\sin x+\tan x -1=\tan x-\sin x$
$\sin x\tan x-1-\tan x+\sin x=0$
$\sin x\tan x-\tan x-1+\sin x=0$
$\tan x(\sin x-1)+1(\sin x-1)=0$
$(\tan x+1)(\sin x-1)=0$
$\Rightarrow \sin x-1=0$ or $\tan x+1=0$
$\sin x=1$ or $\tan x=-1$
$\Rightarrow \sin x=\sin\large\frac{\pi}{2}$ or $\tan x=-1=\tan(-\large\frac{\pi}{4})$
$\Rightarrow x=n\pi+(-1)^n\large\frac{\pi}{2}$
$\therefore$ Required solution is $x=n\pi+(-1)^n \large\frac{\pi}{2}$
Hence (A) is the correct answer.