Prove that \[ \frac{\cos x}{1-\sin x}=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right) \]
Proof:
\[
LHS=\frac{\cos x}{1-\sin x}
\]
Multiply numerator and denominator by
\[
1+\sin x
\]
\[
LHS=\frac{\cos x(1+\sin x)}{(1-\sin x)(1+\sin x)}
\]
Using
\[
(1-\sin x)(1+\sin x)=1-\sin^2x
\]
and
\[
1-\sin^2x=\cos^2x
\]
we get
\[
LHS=\frac{\cos x(1+\sin x)}{\cos^2x}
\]
\[
=\frac{1+\sin x}{\cos x}
\]
Divide numerator and denominator by
\[
\cos x
\]
\[
LHS=\sec x+\tan x
\]
Using the identity:
\[
\sec x+\tan x=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)
\]
Therefore,
\[
LHS=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)
\]
Hence proved,
\[
\boxed{\frac{\cos x}{1-\sin x}=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)}
\]