Question

If exp $\left\{\right(si{n}^{2}x+si{n}^{4}x+si{n}^{6}x+....\infty .\left)lo{g}_{e}2\right\}$ satisfies the equation $x^2-9x+8=0$, find the value of $\frac{cosx}{cosx+sinx},0<x<\pi /2$

• A. $(\sqrt{3}-1)/2$
• B. $(\sqrt{3}-1)/4$
  
• C. $(\sqrt{3}+1)/2$
• D. $(\sqrt{3}-1)/6$

Answer 1

The correct option is A $(\sqrt{3}-1)/2$

$\exp \left\{({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+...\infty )\log 2\right\}=e^{\frac{{\sin }^{2}x}{1-{\sin }^{2}x}\log 2}=e^{\frac{{\sin }^{2}x}{{\cos }^{2}x}\log 2}$

$\Rightarrow 2^{{\tan }^{2}x}$ satisfy $x^2-9x+8=0$
$\Rightarrow x=1,8$

$\therefore 2^{{\tan }^{2}x}=1$ and $\therefore 2^{{\tan }^{2}x}=1$

$\Rightarrow {\tan }^{2}x=0$ and ${\tan }^{2}x=3$

$\Rightarrow x=n\pi$ and $x=n\pi \pm \frac{\pi }{3}$

Neglecting $x=n\pi$ as $0<x<\frac{\pi }{2}$

$\Rightarrow x=\frac{\pi }{3}$

$\therefore \frac{\cos x}{\cos x+\sin x}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\sqrt{3}}{2}}=\frac{1}{1+\sqrt{3}}=\frac{\sqrt{3}-1}{2}$