Answer
Counterexample: $\displaystyle \theta=-\frac{\pi}{2}$
(sample answer)
Work Step by Step
The problem that leads us to search for a counterexample is that
on the RHS, we have a POSITIVE square root of a number.
So, we find a $\theta$ for which $\sin\theta$ is negative.
Sine is negative in quadrants III and IV, so we can select
$ \displaystyle \theta=-\frac{\pi}{2}$ (sample)
LHS$=-1$
RHS=$\sqrt{1-0}=+1$
LHS$\neq$RHS so the identity is not valid.