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