Solve this following


If $e^{\left(\cos ^{2} x+\cos ^{4} x+\cos ^{6} x+\ldots \infty\right) \log _{c} 2}$ satisfies the equation

$\mathrm{t}^{2}-9 \mathrm{t}+8=0$, then the value of

$\frac{2 \sin x}{\sin x+\sqrt{3} \cos x}\left(0


  1. $2 \sqrt{3}$

  2. $\frac{3}{2}$

  3. $\sqrt{3}$

  4. $\frac{1}{2}$

Correct Option: , 4


$e^{\left(\cos ^{2} \theta+\cos ^{4} \theta+\ldots \ldots\right)\left(n^{2}\right.}=2^{\cos ^{2} \theta+\cos ^{4} \theta+\ldots \infty}$

$=2^{\cot ^{2} \theta}$

Now $t^{2}-9 t+9=0 \Rightarrow t=1,8$

$\Rightarrow \quad 2^{\cot ^{2} \theta}=1,8 \Rightarrow \cot ^{2} \theta=0,3$

$0<\theta<\frac{\pi}{2} \Rightarrow \cot \theta=\sqrt{3}$

$\Rightarrow \quad \frac{2 \sin \theta}{\sin \theta+\sqrt{3} \sin \theta}=\frac{2}{1+\sqrt{3} \cot \theta}=\frac{2}{4}=\frac{1}{2}$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now