Given that `cosx/2.cosx/4.cosx/8....=sinx/x` Then find the sum `1/2^2sec^2x/2+1/2^4sec^2x/4+...`

Explanation

Given `\cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} \cdots=\frac{\sin x}{x}`
$$ \begin{aligned} &\therefore \log \left(\cos \frac{x}{2}\right)+\log \left(\cos \frac{x}{4}\right)+\cdots \\ &=\log (\sin x)-\log x \end{aligned} $$
Differentiating twice both sides we get
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