Question

(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x

Answer 1

As

$\tan x=\dfrac{\sin x}{\cos x}$
$\cot x=\dfrac{\cos x}{\sin x}$

we have

$\sin x(\tan x\cos x-\cot x\cos x)=\sin x\left(\sin x-\dfrac{\cos^2x}{\sin x}\right)=\sin^2x-\cos^2x$

Recalling that $\cos2x=\cos^2x-\sin^2x$, we end up with

$-\cos2x=1-2\cos2x$
$1=\cos2x$
$\implies2x=2n\pi$

(where $n$ is any integer)

$\implies x=n\pi$