Lời giải:
Ta có:
\(\frac{1+\sin x}{1-\sin x}+\frac{1-\sin x}{1+\sin x}=\frac{(1+\sin x)^2+(1-\sin x)^2}{(1-\sin x)(1+\sin x)}\)
\(=\frac{1+\sin ^2x+2\sin x+1-2\sin x+\sin ^2x}{1-\sin ^2x}\)
\(=\frac{2(1+\sin ^2x)}{\cos ^2x}=\frac{2(\sin ^2x+\cos ^2x+\sin ^2x)}{\cos ^2x}\)
\(=\frac{4\sin ^2x+2\cos ^2x}{\cos ^2x}=4(\frac{\sin x}{\cos x})^2+2=4\tan ^2x+2=2(1+2\tan ^2x)\)
Ta có đpcm.