Answer
Simplifying the left side using the identity $\sin^{theta} +\sin^2{\theta}=1$ results to $1$, which is equal to the RHS. THus, the given statement is an identity.
Work Step by Step
\[
\begin{aligned}
L H S &: \quad(\sin \theta \cos \phi)^{2}+(\sin \theta \sin \phi)^{2}+\cos^2( \theta )\\
&=\sin ^{2} \theta \cos ^{2} \phi+\sin ^{2} \theta \sin ^{2} \phi+\cos ^{2} \theta
\end{aligned}
\]
Factoring out the common factor $\sin ^{2} \theta \quad$, then using the identity $\sin^{\theta}+\cos^2{\theta}=1$ yields
\[
\begin{array}{l}
=\sin ^{2} \theta\left(\cos^{2} \phi+\sin ^{2} \phi\right)+\cos ^{2} \theta \\
=(\sin ^{2} \theta)(1)+\cos^{2} \theta \\
=\sin ^{2} \theta+\cos^{2} \theta \\
=1
=R H S
\end{array}
\]
Therefore LHS = RHS
Hence the identity is established
