Question

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

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

Answer 1

Answer: (A)

$2+\sqrt{2}$

Answer 2

Here, $\cos \theta+\sin \theta+\frac{2}{\sin 2 \theta}, \theta \in\left(0, \frac{\pi}{2}\right)$
For minimum value, $\sin 2 \theta$ must be maximum.
$\therefore 2 \theta=\frac{\pi}{2}$
$\Rightarrow \theta=\frac{\pi}{4}$
Hence, $\cos \frac{\pi}{4}+\sin \frac{\pi}{4}+\frac{2}{\sin \frac{\pi}{2}}=\sqrt{2}+2$