Question

If $1+cos\alpha +co{s}^2\alpha +...\infty =2-\sqrt{2},then\alpha ,(o<\alpha <\pi )$

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{3\pi }{4}$

Answer 1

The correct option is D $\frac{3\pi }{4}$

Given infinity series,

$$\begin{array}{c}1+\cos a+{\cos }^{2}a+..........\infty =2-\sqrt{2}\\ then,commonratiois-r=\cos a\\ \therefore sumofgeometricinfinityseries....s_{\infty }=\frac{a}{1-r}\\ then,\\ \frac{1}{1-\cos a}=2-\sqrt{2}\\ 1-\cos a=\frac{1}{2-\sqrt{2}}\times \frac{2+\sqrt{2}}{2+\sqrt{2}}=\frac{2+\sqrt{2}}{2}=1+\frac{1}{\sqrt{2}}\\ Now,\\ 1-\cos a=\frac{\sqrt{2}+1}{\sqrt{2}}\Rightarrow \cos a=1-\frac{\sqrt{2}+1}{\sqrt{2}}\\ \cos a=\frac{\sqrt{2}-\sqrt{2}-1}{\sqrt{2}}=-\frac{1}{\sqrt{2}}\\ \therefore \alpha =-\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\\ \alpha =-\frac{\sqrt{2}}{2}=\frac{3\pi }{4}\\ So,thatthecorrectoptionisD.\end{array}$$