Question

If $e ^{\left(\cos ^{2} x+\cos ^{4} x+\cos ^{6} x+\ldots \infty\right) \log _{e} 2}$ satisfies the equation $t ^{2}-9 t +8=0$, 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)

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