Answer
See the explanation below.
Work Step by Step
$\frac{\cos x}{1-\sin x}+\frac{1-\sin x}{\cos x}=2\sec x$
Recall Pythagorean Identity,
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Use the above identity and solve the left side of the given expression,
$\frac{\cos x}{1-\sin x}+\frac{1-\sin x}{\cos x}=2\sec x$
Multiply numerator and denominator of $\frac{\cos x}{1-\sin x}$ by $1+\sin x$.
$\begin{align}
& \frac{\cos x}{1-\sin x}+\frac{1-\sin x}{\cos x}=\frac{\cos x}{1-\sin x}\cdot \left( \frac{1+\sin x}{1+\sin x} \right)+\frac{1-\sin x}{\cos x} \\
& =\frac{\cos x\left( 1+\sin x \right)}{1-{{\sin }^{2}}x}+\frac{1-\sin x}{\cos x} \\
& =\frac{\cos x\left( 1+\sin x \right)}{{{\cos }^{2}}x}+\frac{1-\sin x}{\cos x} \\
& =\frac{1+\sin x}{\cos x}+\frac{1-\sin x}{\cos x}
\end{align}$
Recall Reciprocal Identity,
$\frac{1}{\cos x}=\sec x$
Further simplify,
$\begin{align}
& \frac{\cos x}{1-\sin x}+\frac{1-\sin x}{\cos x}=\frac{1+\sin x+\left( 1-\sin x \right)}{\cos x} \\
& =\frac{2}{\cos x} \\
& =2\sec x
\end{align}$
Therefore,
$\frac{\cos x}{1-\sin x}+\frac{1-\sin x}{\cos x}=2\sec x$
