Answer
The left hand side is equivalent to $1$ therefore the given equation is an identity.
Refer to the solution below.
Work Step by Step
$\text{ LHS } = (1-\cos^2{\theta})(1+\cot^2{\theta})$
$\text{By Expanding:}$
\begin{align}
\text{ LHS } &= 1- \cos^2{\theta}-\cos^2{\theta} \cot^2{\theta} + \cot^2{\theta} \\[3mm]
&= \sin^2{\theta}-\cos^2{\theta} \dfrac{\cos^2{\theta}}{\sin^2{\theta}}+ \dfrac{\cos^2{\theta}}{\sin^2{\theta}} \\[3mm]
&= \sin^2{\theta} - \dfrac{\cos^4{\theta}}{\sin^2{\theta}}+\dfrac{\cos^2{\theta}}{\sin^2{\theta}} \\[3mm]
&= \sin^2{\theta} +\dfrac{\cos^2{\theta}-\cos^4{\theta}}{\sin^2{\theta}} \\[3mm]
&= \sin^2{\theta}+\dfrac{\cos^2{\theta}(1-\cos^2{\theta})}{\sin^2{\theta}} \\[3mm]
&= \sin^2{\theta}+\dfrac{\cos^2{\theta} \sin^2{\theta}}{\sin^2{\theta}} \\[3mm]
&= \sin^2{\theta}+\cos^2{\theta} \\[3mm]
&= 1 \\[3mm]
&= \text{ RHS}
\end{align}
