How to prove this trigonometric identity? Dec 3, 2016

See below

Explanation:

$\cos 2 x = {\cos}^{2} x - {\sin}^{2}$
$\sin 2 x = 2 \sin x \cos x$
Are you happy about these identities?
If so then proving identities, as above, is a matter of doing something and seeing where you get to!!!
LHS can be written

$\frac{\cos x - \left({\cos}^{2} x - {\sin}^{2} x\right) + 2}{3 \sin x - 2 \sin x \cos x}$
=$\frac{\cos x - {\cos}^{2} x + {\sin}^{2} x + 2}{\sin x \left(3 - 2 \cos x\right)}$
RHS $\frac{1 + \cos x}{\sin} x$
Multiply LHS and RHS by $\sin x \left(3 - 2 \cos x\right)$
LHS=$\cos x - {\cos}^{2} x + {\sin}^{2} x + 2$

RHS=$\left(1 + \cos x\right) \left(3 - 2 \cos x\right)$
=$3 - 2 \cos x + 3 \cos x - 2 {\cos}^{2} x$
=$3 + \cos x - 2 {\cos}^{2} x$
Now looking at the LHS and remembering that ${\sin}^{2} x + {\cos}^{2} x = 1$or ${\sin}^{2} x = 1 - {\cos}^{2} x$
LHS=$\cos x - {\cos}^{2} x + \left(1 - {\cos}^{2} x\right) + 2$
=$\cos x - 2 {\cos}^{2} x + 3$

We are there!!!
Difficult for me to write this really clearly here, paper would be easier!!!
BUT THE IMPORTANT THING WITH IDENTITIES IS DO SOMETHING AND YOU WILL GET THERE.