For the first one, we have
[tex]\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)}[/tex]
[tex]=\dfrac{1+\cos x+1-\cos x}{1-\cos^2x}[/tex]
[tex]=\dfrac2{\sin^2x}[/tex]
Multiply this by [tex]\sin x[/tex] and you end up with
[tex]=\dfrac{2\sin x}{\sin^2x}=\dfrac2{\sin x}=2\csc x[/tex]
For the second one,
[tex]-\tan^2x+\sec^2x=-\dfrac{\sin^2x}{\cos^2x}+\dfrac1{\cos^2x}=\dfrac{1-\sin^2x}{\cos^2x}=\dfrac{\cos^2x}{\cos^2x}=1[/tex]
