Find the general solution of each of the following equations:

To Find: General solution.

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

So $\sin x=1=\sin \left(\frac{\pi}{2}\right)$ or $\tan x=-1=\tan \left(\frac{3 \pi}{4}\right)$

Formula used: $\sin \theta=\sin \alpha \Rightarrow \theta=n \pi+(-1)^{n} \alpha, n \in \mid$ and $\tan \theta=\tan \alpha \Rightarrow \theta=k$ $\pi \pm \alpha, k \in l$

$\Rightarrow x=n \pi+(-1)^{n} \frac{\pi}{2}$ or $x=k \pi \pm \frac{3 \pi}{4}$ where $n, k \in I$

So general solution is $x=n \pi+(-1)^{n} \frac{\pi}{2}$ or $x=k \pi \pm \frac{3 \pi}{4}$ where $n, k, \in$ ।