Question

Assertion (A) : If $\sqrt{4{\sin }^4\theta +{\sin }^{2}2\theta }+4{\cos }^2\left(\frac{\pi }{4}-\frac{\theta }{2}\right)=2$ then $\theta$ lies in 3rd quadrant or 4th quadrant .
Reason (R) : $\sqrt{{\sin }^2\theta }=\sin \theta$

• A. Both (A) and (R) are true and (R) is the coll ect explanation of (A)
• B. Both (A) and (R) are true but (R) is not the collect explanation of (A)
• C. (A) is true bur (R) is false
• D. (A) is false but (R) is true

Answer 1

Answer: (C)

Given, $\sqrt{4{\sin }^4\theta +{\sin }^{2}2\theta }+4{\cos }^2\left(\frac{\pi }{4}-\frac{\theta }{2}\right)=2$
$=\sqrt{4{\sin }^4\theta +4{\sin }^2\theta {\cos }^2\theta }+2\left\{1+\cos 2\left(\frac{\pi }{4}-\frac{\theta }{2}\right)\right\}=2$
$=\sqrt{4{\sin }^2\theta \left({\sin }^2\theta +{\cos }^2\theta \right)}+2(1+\sin \theta )=2$
$=\sqrt{4{\sin }^2\theta }+2+2\sin \theta =2$
$\Rightarrow |2\sin \theta |+2\sin \theta =0$
$\Rightarrow \sin \theta <0$
So, $\theta$ lies in $3$ rd or $4$ th quadrant.