Your question seems a bit incomplete, but for starters you can write
[tex]\dfrac1{1-\sin x}-\dfrac{\sin x}{1+\sin x}=\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}-\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}[/tex]
Expanding where necessary, recalling that [tex](1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x[/tex], you have
[tex]\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{1+\sin x-\sin x+\sin^2x}{\cos^2x}=\dfrac{1+\sin^2x}{\cos^2x}[/tex]
and you can stop there, or continue to rewrite in terms of the reciprocal functions,
[tex]\dfrac{1+\sin^2x}{\cos^2x}=\sec^2x+\tan^2x[/tex]
Now, since [tex]1+\tan^2x=\sec^2x[/tex], the final form could also take
[tex]\sec^2x+\tan^2x=\sec^2x+(\sec^2x-1)=2\sec^2x-1[/tex]
or
[tex]\sec^2x+\tan^2x=(1+\tan^2x)+\tan^2x=1+2\tan^2x[/tex]
