The value of frac{1}{4} ,,tan frac{pi}{8 | vedclass

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Question

$\cot x=\frac{1}{2}\left(\cot \frac{x}{2}-	an \frac{x}{2}ight)$

$\cot x=\frac{1}{2}\left\{\frac{1}{2}\left(\cot \frac{x}{4}-	an \frac{x}{4}ig

Vedclass · Accepted Answer

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

Vedclass · Answer

$\cot x=\frac{1}{2}\left(\cot \frac{x}{2}-	an \frac{x}{2}ight)$

$\cot x=\frac{1}{2}\left\{\frac{1}{2}\left(\cot \frac{x}{4}-	an \frac{x}{4}ight)-	an \frac{x}{2}ight\}$

$=\frac{1}{4} \cot \frac{x}{4}-\frac{1}{4} 	an \frac{x}{4}-\frac{1}{2} 	an \frac{x}{2}$

$=\frac{1}{8}\left(\cot \frac{x}{8}-	an \frac{x}{8}ight)-\frac{1}{4} 	an \frac{x}{4}-\frac{1}{2} 	an \frac{x}{2}$

$\Rightarrow \cot x=\frac{1}{2^{n}} \cot \left(\frac{x}{2^{n}}ight)-\frac{1}{2^{n-1}} 	an \left(\frac{x}{2^{n-1}}ight) \dots$

$-\frac{1}{8} 	an \left(\frac{x}{8}ight)-\frac{1}{4} 	an \left(\frac{x}{4}ight)-\frac{1}{2} 	an \left(\frac{x}{2}ight)$

$\Rightarrow \frac{1}{2} 	an \frac{x}{2}+\frac{1}{4} 	an \frac{x}{2}+\frac{1}{8} 	an \frac{x}{8} \ldots \ldots=\frac{1}{2^{n}} \cot \left(\frac{x}{2^{n}}ight)-\cot x$

$\Rightarrow \frac{1}{2} 	an \frac{\mathrm{x}}{2}+\frac{1}{4} 	an \frac{\mathrm{x}}{4}+\ldots . . \infty \mathrm{terms}$

$ = \mathop {\lim }\limits_{n 	o \infty } \,\frac{1}{{{2^n}}}\cot \left( {\frac{x}{{{2^n}}}} ight) - \cot x = \frac{1}{x} - \cot x$

put $x=\frac{\pi}{2}$

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-

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