Question

Prove these two

Answer 1

For the first one, we have

$\dfrac1{1-\cos x}+\dfrac1{1+\cos x}=\dfrac{1+\cos x}{(1-\cos x)(1+\cos x)}+\dfrac{1-\cos x}{(1-\cos x)(1+\cos x)}$

$=\dfrac{1+\cos x+1-\cos x}{1-\cos^2x}$

$=\dfrac2{\sin^2x}$

Multiply this by $\sin x$ and you end up with

$=\dfrac{2\sin x}{\sin^2x}=\dfrac2{\sin x}=2\csc x$

For the second one,

$-\tan^2x+\sec^2x=-\dfrac{\sin^2x}{\cos^2x}+\dfrac1{\cos^2x}=\dfrac{1-\sin^2x}{\cos^2x}=\dfrac{\cos^2x}{\cos^2x}=1$