Question

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

• A. $\frac{\sqrt{3}-1}{2}$
• B. $\frac{\sqrt{3}+1}{2}$
• C. $\frac{\sqrt{3}-1}{4}$
• D. $\frac{\sqrt{3}+1}{4}$

Answer 1

Answer: (A)

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

$\exp \left\{({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+...\infty )\log 2\right\}$

As the given series of $sine$ is in G.P

$\therefore$ $S_{\infty }=\frac{a}{1-r}$

$=e^{\frac{{\sin }^{2}x}{1-{\sin }^{2}x}\log 2}=e^{\frac{{\sin }^{2}x}{{\cos }^{2}x}\log 2}$

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

$\therefore$ $(x-1)(x-8)=0$

$\Rightarrow x=1,8$

$\therefore 2^{{\tan }^{2}x}=1$ 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 $x=n\pi$ $\Rightarrow x=\frac{\pi }{3}$
$\therefore \frac{\cos x}{\cos x+\sin x}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\sqrt{3}}{2}}=\frac{1}{1+\sqrt{3}}=\frac{\sqrt{3}-1}{2}$