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