Question

If $e^{\left[({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+....+\infty ){\log }_{e}2\right]}$ satisfies the equation $x^2-9x+8=0$,then the value of $g\left(x\right)=\frac{\cos x}{\cos x+\sin x}$ is

• A. $\frac{\sqrt{3}+1}{2}$
• B. $\frac{\sqrt{3}-1}{2}$
• C. $8$
• D. None of these

Answer 1

Answer: (B)

$\frac{\sqrt{3}-1}{2}$

Consider, $y=exp\left[({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+....+\infty ){\log }_{e}2\right]$
$\Rightarrow y=exp\left[\left(\frac{{\sin }^{2}x}{1-{\sin }^{2}x}\right){\log }_{e}2\right]=exp\left[{\tan }^{2}x{\log }_{e}2\right]=2^{{\tan }^{2}x}$
Since, $y$ satisfies $x^2-9x+8=0$, then
$y=1,8$
$\Rightarrow 2^{{\tan }^{2}x}=2^0,2^3$
$\Rightarrow {\tan }^{2}x=0,3$
$\Rightarrow \tan x=0,\pm \sqrt{3}$
Now, $g\left(x\right)=\frac{\cos x}{\cos x+\sin x}=\frac{1}{1+\tan x}=1,\frac{1}{1\pm \sqrt{3}}=1,\frac{-\sqrt{3}-1}{2},\frac{\sqrt{3}-1}{2}$
Ans: B