Question

If $\theta \in (\frac{\pi }{2},\frac{3\pi }{2})$, 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. $-2\cot \theta$
• B. $2\cot \theta$
• C. $2\cos \theta$
• D. $2\sin \theta$

Answer 1

The correct option is B $2\cot \theta$

$\sqrt{4{\cos }^4\theta +{\sin }^{2}2\theta }+4\cot \theta {\cos }^2(\frac{\pi }{4}-\frac{\theta }{2})$
Using $\cos 2x=2{\cos }^{2}x-1$, we get
$\sqrt{4{\cos }^4\theta +{\sin }^{2}2\theta }+2\cot \theta (1+\cos (\frac{\pi }{2}-\theta ))=\sqrt{4{\cos }^4\theta +(2\sin \theta \cos \theta {)}^2}+2\cot \theta (1+\sin \theta )=|2\cos \theta |\sqrt{{\cos }^2\theta +{\sin }^2\theta }+2\cot \theta (1+\sin \theta )=|2\cos \theta |+2\cot \theta (1+\sin \theta )=|2\cos \theta |+2\cot \theta +2\cos \theta \because \theta \in (\frac{\pi }{2},\frac{3\pi }{2})\Rightarrow |2\cos \theta |=-2\cos \theta$
So, the expression $|2\cos \theta |+2\cot \theta +2\cos \theta =2\cot \theta$