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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-
$\frac{5}{\pi}-\frac{1}{2}$
$\frac{3}{\pi}+\frac{1}{2}$
$\frac{2}{\pi}-\frac{1}{2}$
$\frac{4}{\pi}-\frac{1}{4}$
Solution
$\cot x=\frac{1}{2}\left(\cot \frac{x}{2}-\tan \frac{x}{2}\right)$
$\cot x=\frac{1}{2}\left\{\frac{1}{2}\left(\cot \frac{x}{4}-\tan \frac{x}{4}\right)-\tan \frac{x}{2}\right\}$
$=\frac{1}{4} \cot \frac{x}{4}-\frac{1}{4} \tan \frac{x}{4}-\frac{1}{2} \tan \frac{x}{2}$
$=\frac{1}{8}\left(\cot \frac{x}{8}-\tan \frac{x}{8}\right)-\frac{1}{4} \tan \frac{x}{4}-\frac{1}{2} \tan \frac{x}{2}$
$\Rightarrow \cot x=\frac{1}{2^{n}} \cot \left(\frac{x}{2^{n}}\right)-\frac{1}{2^{n-1}} \tan \left(\frac{x}{2^{n-1}}\right) \dots$
$-\frac{1}{8} \tan \left(\frac{x}{8}\right)-\frac{1}{4} \tan \left(\frac{x}{4}\right)-\frac{1}{2} \tan \left(\frac{x}{2}\right)$
$\Rightarrow \frac{1}{2} \tan \frac{x}{2}+\frac{1}{4} \tan \frac{x}{2}+\frac{1}{8} \tan \frac{x}{8} \ldots \ldots=\frac{1}{2^{n}} \cot \left(\frac{x}{2^{n}}\right)-\cot x$
$\Rightarrow \frac{1}{2} \tan \frac{\mathrm{x}}{2}+\frac{1}{4} \tan \frac{\mathrm{x}}{4}+\ldots . . \infty \mathrm{terms}$
$ = \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{{{2^n}}}\cot \left( {\frac{x}{{{2^n}}}} \right) – \cot x = \frac{1}{x} – \cot x$
put $x=\frac{\pi}{2}$
