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