Prove the identity
\(\displaystyle \frac{1+\sin x}{1-\sin x}\equiv (\tan x+\sec x)^2.\)
Solution
\[
(\tan x+\sec x)^2=\left(\frac{\sin x}{\cos x}+\frac{1}{\cos x}\right)^2
=\frac{(1+\sin x)^2}{\cos^2x}
=\frac{(1+\sin x)^2}{1-\sin^2x}
=\frac{1+\sin x}{1-\sin x}.
\]
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