Question

If $\frac{\pi }{2}<\theta <\frac{3\pi }{2}$, then the value of $=\sqrt{4{\cos }^4\theta +{\sin }^{2}2\theta }+4\cot \theta {\cos }^2(\frac{\pi }{4}-\frac{\theta }{2})$ is

• A. $2\sin \theta$
• B. $-2\sin \theta$
• C. $2\cot \theta$
• D. $-2\cot \theta$

Answer 1

Answer: (C)

$2\cot \theta$

Given expression $\sqrt{4{\cos }^4\theta +{\sin }^{2}2\theta }+4\cot \theta {\cos }^2(\frac{\pi }{4}-\frac{\theta }{2})$

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

$=\left|2\cos \theta \right|+2\cot \theta (1+\sin \theta )$

$=-2\cos \theta +2\cot \theta +2\cos \theta$

$[\because \frac{\pi }{2}<\theta <\frac{3\pi }{2},\therefore \cos \theta$ is negative $]$

$=2\cot \theta$