Question

If $\cos \frac{x}{2}\cos \frac{x}{2^2}\cos \frac{x}{2^3}\dots \cos \frac{x^{}}{2^n}=f\left(x\right)$ , then $\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+\dots +\frac{1}{2^n}\tan \frac{x}{2^n}=$

• A. $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$
• B. $\frac{f\left(x\right)}{f^{\prime }\left(x\right)}$
• C. $\frac{-f^{\prime }\left(x\right)}{f\left(x\right)}$
• D. $0$

Answer 1

The correct option is A $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$

$\cos \frac{x}{2}\cos \frac{x}{2^2}\cos \frac{x}{2^3}....\cos \frac{x}{2^n}=f\left(x\right)$

$\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$

$\sin x=2^2\sin \frac{x}{4}\cos \frac{x}{4}\cos \frac{x}{2}$

$\sin x=2^3\sin \frac{x}{8}\cos \frac{x}{8}\cos \frac{x}{4}\cos \frac{x}{2}$

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$\sin x=2^n\sin \frac{x}{2^n}{\prod }_{k=1}^n\cos \frac{x}{2^k}$

Also

$\cos \left(\frac{x}{2}\right)\cos \left(\frac{x}{4}\right)\cos \left(\frac{x}{8}\right)=\frac{\left(2\cos \left(\frac{x}{2}\right)\sin \left(\frac{x}{2}\right)\right)\cos \left(\frac{x}{4}\right)\cos \left(\frac{x}{8}\right)}{2\sin \frac{x}{2}}$

$=\frac{\sin x\left(\cos \left(\frac{x}{4}\right)\sin \left(\frac{x}{4}\right)\right)\cos \left(\frac{x}{8}\right)}{\left(2\sin \left(\frac{x}{2}\right).2\sin \left(\frac{x}{2}\right)\right)}=\frac{\sin x\cos \left(\frac{x}{8}\right)}{4\sin \left(\frac{x}{4}\right)}$

$=\frac{\sin x\left(2\sin \left(\frac{x}{8}\right)\cos \left(\frac{x}{8}\right)\right)}{8\sin \left(\frac{x}{4}\right)\sin \left(\frac{x}{8}\right)}$

$\cos \left(\frac{x}{2}\right)\cos \left(\frac{x}{4}\right)\cos \left(\frac{x}{8}\right)=\frac{\sin x}{8\sin \left(\frac{x}{8}\right)}$

$\cos \left(\frac{x}{2}\right)\cos \left(\frac{x}{2^2}\right)....\cos \left(\frac{x}{2^n}\right)=\frac{\sin x}{2^n\sin \left(\frac{x}{2^n}\right)}=f\left(x\right)$

$n\to \infty ,2^n\to \infty ,\frac{x}{2^n}\to 0,\frac{\sin \left(\frac{x}{2^n}\right)}{\frac{x}{2^n}}....\left(1\right)$

$\cos \left(\frac{x}{2}\right)\cos \left(\frac{x}{2^2}\right)....\cos \left(\frac{x}{2^n}\right)=\frac{\sin x}{x}=f\left(x\right)$

$\sin \left(\frac{x}{2^{}}\right)=\frac{1}{2}\cos \left(\frac{x}{2}\right)\Rightarrow {\sum }_{k=1}^{kn=}\frac{1}{2^k}\tan \left(\frac{k}{2^k}\right)=\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$

$\frac{1}{2^{}}\tan \frac{x}{2^{}}+\frac{1}{2^2}\tan \frac{x}{2^2}+....\frac{1}{2^n}\tan \frac{x}{2^n}=\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$

option (A) is correct