Assertion given:-

$\sqrt{4{\sin }^4\theta +{\sin }^{2}2\theta }+4{\cos }^2\left(\frac{\pi }{2}-\frac{\theta }{2}\right)=2$

Simplify the above expression.

$\sqrt{4{\sin }^4\theta +4{\sin }^2\theta {\cos }^2\theta }+2\left\{1+\cos 2\left(\frac{\pi }{2}-\frac{\theta }{2}\right)\right\}=2$

$\Rightarrow \sqrt{4{\sin }^2\theta \left({\sin }^2\theta +{\cos }^2\theta \right)}+2\{1+\sin \theta \}=2$

$\Rightarrow \sqrt{4{\sin }^2\theta }+2+2\sin \theta =2$

$\Rightarrow \left|2\sin \theta \right|+2\sin \theta =0$

$\Rightarrow \sin \theta <0$

Therefore, $\theta$ will be lying in the third or fourth quadrant.

So, the given assertion is true. 

Reason given:- $\sqrt{{\sin }^2\theta }=\sin \theta$ 

$\because x^2=a^2\Rightarrow x=\pm a$

So, the given reason is false.