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.