Question

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

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

Answer 1

The correct option is C $\frac{1}{2}$

$e^{({\cos }^{2}x+{\cos }^4+\cdots \infty )\ln 2}=2^{{\cos }^{2}x+{\cos }^{4}x+\cdots \infty }$
$=2^{\frac{{\cos }^{2}x}{1-{\cos }^{2}x}}$ (sum of infinite G.P.)
$=2^{{\cot }^{2}x}$

$t^2-9t+8=0\Rightarrow t=1,8$
$\Rightarrow 2^{{\cot }^{2}x}=1,8\Rightarrow x=0,3$
$0<x<\frac{\pi }{2}\Rightarrow \cot x=\sqrt{3}$
Hence
$\frac{2\sin x}{\sin x+\sqrt{3}\cos x}=\frac{2}{1+\sqrt{3}\cot x}=\frac{1}{2}$