Question

Given that $\cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}\cdots =\frac{\sin x}{x}$, then expression $\frac{1}{2^2}{\sec }^2\frac{x}{2}+\frac{1}{2^4}{\sec }^2\frac{x}{4}+\cdots$ equals

• A. ${cosec}^{2}x-\frac{1}{x^2}$
• B. $\frac{1}{x^2}$
• C. ${cosec}^{2}x$
• D. None of these

Answer 1

The correct option is A ${cosec}^{2}x-\frac{1}{x^2}$

Given $\cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}\cdots =\frac{\sin x}{x}$
$\therefore \log \left(\cos \frac{x}{2}\right)+\log \left(\cos \frac{x}{4}\right)+\cdots$
$=\log \left(\sin x\right)-\log x$
Differentiating twice both sides we get
$\frac{1}{2^2}{\sec }^2\frac{x}{2}+\frac{1}{2^4}{\sec }^2\frac{x}{4}+\cdots ={cosec}^{2}x-\frac{1}{x^2}$