If  $exp \left\{\left({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+\mathrm{...}\text{upto\hspace{0.17em}}\infty \right)\ln 2\right\}$ 

satisfies the equation $x^2-17x+16=0$  then  the 

value of  $\frac{2\cos x}{\sin x+2\cos x}\left(0<x<\pi /2\right)$ is 

We have  ${\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+\mathrm{...}$
$=\frac{{\sin }^2}{1-{\sin }^{2}x}={\tan }^{2}x$
Therefore,  $=\exp \left\{\left({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+\mathrm{...}\text{upto}\infty \right)\ln 2\right\}$$=\exp \left\{{\tan }^{2}x\ln 2\right\}=\exp \left\{\ln 2^{{\tan }^{2}x}\right\}$$=2^{{\tan }^{2}x}$
 As  $\alpha$ satisfies the equation $x^2-17x+16=0$  we get 
Since $0<x<\pi /2,{\tan }^{2}x>0\Rightarrow \alpha =2^{{\tan }^{2}x}>1.$ Therefore,$2^{{\tan }^{2}x}=16=2^4\Rightarrow {\tan }^{2}x=4\Rightarrow \tan x=2\left[\because \tan x>0\right]$
Thus,  $\frac{2\cos x}{\sin x+2\cos x}=\frac{2}{\tan x+2}=\frac{2}{2+2}=\frac{1}{2}$