Question

If $y=\frac{2}{\sin \theta +\sqrt{3}\cos \theta },$ then the minimum value of y($\theta \in [0,\frac{\pi }{2}]$) is

• A. $1$
• B. $2$
• C. $\frac{1}{\sqrt{3}+1}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $1$

$Y=\frac{2}{\sin \theta +\sqrt{3}\cos \theta }$

$Y\to min;$

$\sin \theta +\sqrt{3}\cos \theta \to max$

$G\left(\theta \right)=\cos \theta -\sqrt{3}\sin \theta$

$\tan \theta =\frac{1}{\sqrt{3}};$

$\theta ={30}^{\circ }$

$G\left({30}^{\circ }\right)=\frac{1}{2}+\sqrt{3}.\frac{\sqrt{3}}{2}=2$

$Y_{min}=\frac{2}{G\left({30}^{\circ }\right)}=\frac{2}{2}=1$