Question

The minimum value of $\cos \theta +\sin \theta +\frac{2}{\sin 2\theta }$ for $\theta \in (0,\frac{\pi }{2})$, is

• A. $2+\sqrt{2}$
• B. $2$
• C. $1+\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

The correct option is A $2+\sqrt{2}$

Here,

$\cos \theta +\sin \theta +\frac{2}{\sin 2\theta }\theta ,\in (0,\frac{\pi }{2})$

For minimum value,

$\sin 2\theta$ must be maximum.

$\therefore 2\theta =\frac{\pi }{2}\Rightarrow \theta =\frac{\pi }{4}$
Hence,

$\cos \frac{\pi }{4}+\sin \frac{\pi }{4}+\frac{2}{\sin \frac{\pi }{2}}=\sqrt{2}+2$