Answer
See the explanation below.
Work Step by Step
Let us consider the right side of the given expression:
$\frac{{{\sin }^{2}}\alpha +1-{{\cos }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)}$
By using the trigonometric identity
$\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}$ and ${{\sin }^{2}}x=1-{{\cos }^{2}}x$,
The above expression can be further simplified as:
$\begin{align}
& \frac{{{\sin }^{2}}\alpha +1-{{\cos }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)}=\frac{{{\sin }^{2}}\alpha +{{\sin }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)} \\
& =\frac{2{{\sin }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)} \\
& =2\left( \frac{\sin \alpha }{1+\cos \alpha } \right) \\
& =2\tan \frac{\alpha }{2}
\end{align}$
Hence, the left side of the given expression is equal to the right side, which is $2\tan \frac{\alpha }{2}=\frac{{{\sin }^{2}}\alpha +1-{{\cos }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)}$.
