Question

If exp $\left[\left(\sin^{2} x+\sin ^{4} x+\sin ^{6} x+\ldots \infty\right) \ln 2\right]$ satisfies the equation $y^{2}-9 y+8=0$, 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

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 $y=1=e^{\tan^{2} x \cdot \ln 2}$
$\Rightarrow e^{0}=e^{\tan^{2} x \cdot \ln 2}$
$\Rightarrow 0=\tan^{2}x\cdot \ln2$
$\Rightarrow \tan^{2}x=0,$ 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, $\frac{\cos x}{\cos x + \sin x}$
$=\frac{\frac{1}{2}}{\frac{1}{2} + \frac{\sqrt{3}}{2}}$
$=\frac{1}{\sqrt{3} + 1}=\frac{\sqrt{3} - 1}{2}$
$\therefore \frac{\sqrt{a} - 1}{2}=\frac{\sqrt{3} - 1}{2}\Rightarrow a=3$