If $\cos\frac{x}{2}.\cos\frac{x}{2^{2}}\ldots..\cos\frac{x}{2^{n}} = \frac{\sin x}{2^{n}\sin{\ \frac{x}{2^{n}}}}$ , then $\frac{1}{2} an{\frac{x}{2} + \frac{1}{2^{2}} an{\frac{x}{2^{2}} + ... + \frac{1}{2^{n}} an\frac{x}{2^{n}}}}$ is

Question

If $\cos\frac{x}{2}.\cos\frac{x}{2^{2}}\ldots..\cos\frac{x}{2^{n}} = \frac{\sin x}{2^{n}\sin{\ \frac{x}{2^{n}}}}$ , then $\frac{1}{2}	an{\frac{x}{2} + \frac{1}{2^{2}}	an{\frac{x}{2^{2}} + ... + \frac{1}{2^{n}}	an\frac{x}{2^{n}}}}$ is

Dr. Abubacker · Accepted Answer

(b)
$\frac{1}{2}	an{\frac{x}{2} = \frac{1}{2}\cot{\frac{x}{2} - \cot{x\ \ \ \ \ \ \ \ \left\lbrack \because\cot x = \frac{1 - 	an^{2}\frac{x}{2}}{2	an\frac{x}{2}} ightbrack}}}$
And $\frac{1}{2^{2}}	an{\frac{x}{2^{2}} = \frac{1}{2^{2}}\cot\left( \frac{x}{2^{2}} ight)} - \frac{1}{2}\cot\left( \frac{x}{2} ight)$
Similarly, $\frac{1}{2^{3}}	an\left( \frac{x}{2^{3}} ight) = \frac{1}{2^{3}}\cot\left( \frac{x}{2^{3}} ight) - \frac{1}{2^{2}}\cot{\ldots\left( \frac{x}{2^{2}} ight)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \vdots \ \ \ \ \  \vdots \ \ \ \ \ \ \ \ \  \vdots \ \ \ \ \  \vdots \ \ \ \ \ \ \ \ \  \vdots$
$\frac{1}{2^{n}}	an{\left( \frac{x}{2^{n}} ight) = \frac{1}{2^{n}}\cot\left( \frac{x}{2^{n}} ight) - \frac{1}{2^{n - 1}}\cot\left( \frac{x}{2^{n - 1}} ight)}$
Om adding all the above results, we get
$\frac{1}{2}	an{\frac{x}{2} + \frac{1}{2^{2}}	an{\left( \frac{x}{2^{2}} ight) + ... + \frac{1}{2^{n}}	an\left( \frac{x}{2^{n}} ight) = \frac{1}{2^{n}}\cot{\left( \frac{x}{2^{n}} ight) - \cot x}}}$

Q. If $\cos\frac{x}{2}.\cos\frac{x}{2^{2}}\ldots..\cos\frac{x}{2^{n}} = \frac{\sin x}{2^{n}\sin{\ \frac{x}{2^{n}}}}$ , then $\frac{1}{2}\tan{\frac{x}{2} + \frac{1}{2^{2}}\tan{\frac{x}{2^{2}} + ... + \frac{1}{2^{n}}\tan\frac{x}{2^{n}}}}$ is

Explanation

Reviewed by Dr. Abubacker

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