Question

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

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

Answer 1

The correct option is A $\frac{\sqrt{3}-1}{2}$

$\text{exp}\left[\right({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+....\infty \left){\log }_{e}2\right]=e^{\frac{{\sin }^{2}x}{1-{\sin }^{2}x}\cdot {\log }_{e}2}=e^{{\log }_{e}2\frac{{\sin }^{2}x}{{\cos }^{2}x}}\Rightarrow 2^{{\tan }^{2}x}\text{satisfies}{x}^2-9x+8=0\Rightarrow x=1,8\therefore 2^{{\tan }^{2}x}=1\text{and}{2}^{{\tan }^{2}x}=8\Rightarrow {\tan }^{2}x=0\text{and}{\tan }^{2}x=3\Rightarrow x=n\pi \text{and}{\tan }^{2}x={\left(\tan \frac{\pi }{3}\right)}^2\Rightarrow x=n\pi \text{and}x=n\pi \pm \frac{\pi }{3}\text{Neglecting}x=n\pi \text{as}0<x<\frac{\pi }{2}\Rightarrow x=\frac{\pi }{3}\in (0,\frac{\pi }{2})\therefore \frac{\cos x}{\cos x+\sin x}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{\sqrt{3}}{2}}=\frac{1}{1+\sqrt{3}}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}\Rightarrow \frac{\cos x}{\cos x+\sin x}=\frac{\sqrt{3}-1}{2}$