Question

If exp $\left[\left({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+\dots \infty \right)\ln 2\right]$ satisfies the equation $y^2-9y+8=0$, and the value of $\frac{\cos x}{\cos x+\sin x},0<x<\frac{\pi }{2}$, is $\frac{\sqrt{a}-1}{2}$ then find $a$.

Answer 1

Answer: 3

We have, $e^{\left[\left({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+...\infty \right)\ln 2\right]}$
$=e^{\left(\frac{{\sin }^{2}x}{1-{\sin }^{2}x}\right)\ln 2}=e^{{\tan }^{2}x\cdot \ln 2}$
$\therefore exp\left[\left({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+...\infty \right)\ln 2\right]=e^{{\tan }^{2}x\cdot \ln 2}$
It satisfies the given equation is $y^2-9y+8$ .
$\left(y-1\right)\left(y-8\right)=0$
$\Rightarrow y=1,8$
If $y=1=e^{{\tan }^{2}x\cdot \ln 2}$
$\Rightarrow e^0=e^{{\tan }^{2}x\cdot \ln 2}$
$\Rightarrow 0={\tan }^{2}x\cdot \ln 2$
$\Rightarrow {\tan }^{2}x=0,$ but $x\in \left(0,\frac{\pi }{2}\right),\therefore$ neglecting $x=0$
If $y=8=e^{{\tan }^{2}x\cdot \ln 2}$
$\Rightarrow 2^3=e^{{\tan }^{2}x\cdot \ln 2}$
Taking $\ln$ both sides we get,
$3={\tan }^{2}x\Rightarrow \tan x=\pm \sqrt{3}$
$\Rightarrow \tan x=\tan \left(\pm \frac{\pi }{3}\right)$
$\therefore x=\frac{\pi }{3}$ as, $0<x<\frac{\pi }{2}$
Then, $\frac{\cos x}{\cos x+\sin x}$
$=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\sqrt{3}}{2}}$
$=\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}$
$\therefore \frac{\sqrt{a}-1}{2}=\frac{\sqrt{3}-1}{2}\Rightarrow a=3$