Lời giải:
a.
\(\frac{1}{1-\sin x}+\frac{1}{1+\sin x}=\frac{1+\sin x+1-\sin x}{(1+\sin x)(1-\sin x)}=\frac{2}{1-\sin ^2x}=\frac{2}{\cos ^2x}\)
b.
\(\frac{\tan ^2x-\sin ^2x}{\sin ^2x}=\frac{\frac{\sin ^2x}{\cos ^2x}-\sin ^2x}{\sin ^2x}\)
\(=\frac{1}{\cos ^2x}-1\)