Question

The value of $\frac{1}{4}\tan \frac{\pi }{8}+\frac{1}{8}\tan \frac{\pi }{16}+\frac{1}{16}\tan \frac{\pi }{32}+......\infty$ terms is equal to -

• A. $\frac{5}{\pi }-\frac{1}{2}$
• B. $\frac{3}{\pi }+\frac{1}{2}$
• C. $\frac{2}{\pi }-\frac{1}{2}$
• D. $\frac{4}{\pi }-\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{2}{\pi }-\frac{1}{2}$

Let $S=\frac{1}{4}\tan \frac{\pi }{8}+\frac{1}{8}\tan \frac{\pi }{16}+\frac{1}{16}\tan \frac{\pi }{32}+.....\infty$

$=\frac{1}{2}\{\frac{1}{2}\tan \left(\frac{\pi /4}{2}\right)+\frac{1}{4}\tan \left(\frac{\pi /4}{4}\right)+\frac{1}{8}\tan \left(\frac{\pi /4}{8}\right)+.\dots .\infty \}$

Let $x=\pi /4$

$\therefore s=\frac{1}{2}\{\frac{1}{2}\tan \left(\frac{x}{2}\right)+\frac{1}{2^2}\tan \left(\frac{x}{2^2}\right)+\frac{1}{2^3}\tan \left(\frac{x}{2^3}\right)+....\infty \}$

$=\frac{1}{2}\sum _{n=1}^{\infty }\left(\frac{1}{2^n}\tan \frac{x}{2^n}\right)$

$=\frac{1}{2}(\frac{1}{x}-\cot x)$ $[\because \sum _{n=1}^{\infty }\frac{1}{2^n}\tan \frac{x}{2^n}=\frac{1}{x}-\cot x]$

$=\frac{1}{2}(\frac{1}{\pi /4}-\cot \pi /4)$

$=\frac{1}{2}\times \frac{\pi }{4}-\frac{1}{2}\times \cot \pi /4$

$=\frac{4}{2\pi }-\frac{1}{2}$

$=\frac{2}{\pi }-\frac{1}{2}$.