Answer
The left hand side is equivalent to $2$ therefore the given equation is an identity.
Refer to the solution below.
Work Step by Step
$\text{ LHS } = (\sin{\theta}+\cos{\theta})^2+(\sin{\theta}-\cos{\theta})^2$
$\text{By Expanding:}$
\begin{align}
\text{LHS } &= (\sin^2{\theta}+2 \sin{\theta} \cos{\theta} + \cos^2{\theta})+(\sin^2{\theta}-2 \sin{\theta} \cos{\theta} + \cos^2{\theta}) \\[2mm]
&= \sin^2{\theta}+\cos^2{\theta}+\sin^2{\theta}+\cos^2{\theta} \\[2mm]
&= 2\sin^2{\theta}+2 \cos^2{\theta} \\[2mm]
& = 2(\sin^2{\theta}+\cos^2{\theta}) \\[2mm]
&= 2(1)\\[2mm]
&= 2\\[2mm]
&= \text{ RHS}
\end{align}
