Question

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

• A. $3-\sqrt{2}$
• B. $2+\sqrt{3}$
• C. $2-\sqrt{3}$
• D. $3+\sqrt{2}$

Answer 1

The correct option is C $2-\sqrt{3}$

$\because 0<x<\pi /2$
$\therefore 0<{\sin }^{2}x<1$
Then ${\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+...+\infty =\frac{{\sin }^{2}x}{1-{\sin }^{2}x}={\tan }^{2}x$ .... As series is in G.P.
$\therefore exp\left[\right({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+...+\infty \left){\log }_{e}2\right]=exp\left[{\tan }^{2}x{\log }_{e}2\right]$
$=exp\left\{{\log }_e{2}^{{\tan }^{2}x}\right\}=e^{{\log }_e{2}^{{\tan }^{2}x}}$
$=2^{{\tan }^{2}x}$
Let $y=2^{{\tan }^{2}x}$
$\because y$ satisfies quadratic equation
$y^2-9y+8=0$
$\therefore y=1,8$
if $y=1=2^{{\tan }^{2}x}$
$\therefore 2^{{\tan }^{2}x}=2^0$
$\therefore {\tan }^{2}x=0$
$\therefore x=0$ (impossible) $\because x>0$
Now if $y=8=2^{{\tan }^{2}x}$
$\therefore 2^{{\tan }^{2}x}=2^3$
$\Rightarrow {\tan }^{2}x=3$
$\therefore \tan x=\sqrt{3}$
$\therefore \frac{\sin x-\cos x}{\sin x+\cos x}=\frac{\tan x-1}{\sin x+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{(\sqrt{3}-1{)}^2}{3-1}$
$=\frac{3+1-2\sqrt{3}}{2}$
$=2-\sqrt{3}$