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How to solve this one I know this is related to differentiation but how to proceed with this??? Please give all steps so that it is easily understood.

B. Goddard
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Rahul Laddha
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Taking logs of both sides, $$ \sum_{k=1}^{\infty} \log{\cos{\left( \frac{x}{2^k} \right)}} = \log{\sin{x}}-\log{x}. $$ Differentiating twice, $$ \begin{align} \sum_{k=1}^{\infty} -\frac{1}{2^k} \tan{\left(\frac{x}{2^k} \right)} &= \cot{x}-\frac{1}{x} \\ \sum_{k=1}^{\infty} -\frac{1}{2^{2k}} \sec^2{\left(\frac{x}{2^k} \right)} &= -\csc^2{x}+\frac{1}{x^2}. \end{align} $$

Chappers
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    Good idea! But what if you have some $x$ such that $\cos (x/2^k)<0$ for some $k$? – Arnaldo Jan 03 '17 at 13:44
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    The simplest thing to do for real $x$ would be to square both sides, then take logs and divide by $2$. The derivatives will be the same. For complex arguments, proving it on a neighbourhood of zero and applying meromorphic continuation is enough. – Chappers Jan 03 '17 at 14:02