Question

If $e^{\left({\cos }^{2}x+{\cos }^{4}x+{\cos }^{6}x+\dots \infty \right){\log }_{e}2}$ satisfies the equation $t^2-9t+8=0$, then the value of $\frac{2\sin x}{\sin x+\sqrt{3}\cos x}\left(0<x<\frac{\pi }{2}\right)$ is

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

Answer 1

Answer: (D)

$e^{\left({\cos }^2\theta +{\cos }^4\theta +\dots \infty \right)\ell n^2}=2^{{\cos }^2\theta +{\cos }^4\theta +\dots \infty }$
$=2^{{\cot }^2\theta }$
Now $t^2-9t+9=0$
$\Rightarrow t=1,8$
$\Rightarrow 2^{{\cot }^2\theta }=1,8$
$\Rightarrow {\cot }^2\theta =0,3$
$0<\theta <\frac{\pi }{2}$
$\Rightarrow \cot \theta =\sqrt{3}$
$\Rightarrow \frac{2\sin \theta }{\sin \theta +\sqrt{3}\sin \theta }$
$=\frac{2}{1+\sqrt{3}\cot \theta }$
$=\frac{2}{4}=\frac{1}{2}$