Question

Solve: $\sin x\tan x-1=\tan x-\sin x$.

Answer 1

sinx tanx-1 = tanx - sinx

$sinx\frac{sinx}{cosx}-1=\frac{sinx}{cosx}-sinx$

$\frac{si{n}^{2}x-cosx}{cosx}=\frac{sinx[1-cosx]}{cosx}$

$\Rightarrow$$si{n}^{2}x-cosx=sinx-sinxcosx$

$\Rightarrow$$si{n}^2-sinx-cosx+sinxcosx=0$

$\Rightarrow$$sinx[sinx-1]+cosx[sinx-1]=0$

$\Rightarrow$$[sinx+cosx][sinx-1]=0$

$\Rightarrow [sinx+cosx=0]$ or $[sinx=1]$

then $x=(2n+1)\frac{\pi }{2}$