If exp $\left({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+\dots \mathrm{\infty }\right)$ In 2 satisfies the equation $x^2-9x+8=0$, , then the value of $\frac{\cos x}{\cos x+\sin x},0<x<\frac{\pi }{2}$

We have, 

$\begin{array}{l}{\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+\dots \\ =\frac{{\sin }^{2}x}{1-{\sin }^{2}x}\\ ={\tan }^{2}x\end{array}$

Now,  $x^2-9x+8=0$

Now, $x^2-9x+8=0$

$\begin{array}{l}\Rightarrow (x-1)(x-8)=0\\ \Rightarrow x=1,8\end{array}$

when $2^{{\tan }^{2}x}=1=2^0$

$\Rightarrow {\tan }^{2}x=0$

$\Rightarrow \tan x=0$

when  $2^{{\tan }^{2}x}=8=2^3$

$\begin{array}{l}\Rightarrow {\tan }^{2}x=3\\ \Rightarrow \tan x=\sqrt{3}\end{array}$

Thus, $\frac{\cos x}{\cos x+\sin x}=\frac{1}{1+\tan x}$

$\begin{array}{l}=\frac{1}{1+\sqrt{3}}\\ =\left(\frac{\sqrt{3}-1}{2}\right)\end{array}$