Answer
See the explanation below.
Work Step by Step
By using identities:
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, $2\sin x\cos x=\sin 2x$ and ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
Now, the left-hand side can be written as:
$\begin{align}
& {{\left( \sin \frac{x}{2}+\cos \frac{x}{2} \right)}^{2}}={{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}+{{\cos }^{2}}\frac{x}{2} \\
& =\left[ 2\sin \frac{x}{2}\cos \frac{x}{2} \right]+\left[ {{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2} \right] \\
& =\sin \left( 2\times \frac{x}{2} \right)+1 \\
& =\sin x+1
\end{align}$
Thus, the left side of the equation is equal to $\sin x+1$.
Hence, the left-hand side is equal to $\sin x+1$.
