Question

The value of $\sum _{n=1}^{\infty }\frac{\tan \left(\frac{\theta }{2^n}\right)}{2^{n-1}\cos \frac{\theta }{2^{n-1}}}$

• A. $\frac{2}{\sin 2\theta }-\frac{1}{\theta }$
• B. $\frac{2}{\sin 2\theta }+\frac{1}{\theta }$
• C. $\frac{1}{\sin 2\theta }-\frac{1}{\theta }$
• D. $\frac{1}{\sin \theta }-\frac{1}{\theta }$

Answer 1

The correct option is A $\frac{2}{\sin 2\theta }-\frac{1}{\theta }$

$\begin{array}{c}{\sum }_{n=1}^{\infty }\frac{\tan \left(\frac{\theta }{2^n}\right)}{2^{n-1}\cos \frac{\theta }{2^{n-1}}}\\ \frac{\tan \left(\frac{\theta }{2^n}\right)}{2^{n-1}\cdot \cos \frac{\theta }{2^{n-1}}}=\frac{\sin \frac{\theta }{2^n}}{2^{n-1}\cos \frac{\theta }{2^n}\cdot \cos \frac{\theta }{2^{n-1}}}\\ =\frac{2{\sin }^2\frac{\theta }{2^n}}{2^{n-1}\sin \frac{\theta }{2^{n-1}}\cdot \cos \frac{\theta }{2^{n-1}}}\\ =\frac{1-\cos \frac{\theta }{2^{n-1}}}{2^{n-2}\cdot \sin \frac{\theta }{2^{n-2}}}\\ =\frac{1}{2^{n-2}}\left\{\frac{1}{\sin \frac{\theta }{2^{n-2}}}-\frac{\cos \frac{\theta }{2^{n-2}}}{2\sin \frac{\theta }{2^{n-1}}\cdot \cos \frac{\theta }{2^{n-1}}}\right\}\\ =\frac{1}{2^{n-2}}\left\{\frac{1}{\sin \frac{\theta }{2^{n-2}}}-\frac{1}{2\sin \frac{\theta }{2^{n-1}}}\right\}\\ Now,\\ {\sum }_{n=1}^{{}^{\infty }}\left(\frac{1}{2^{n-2}\cdot \sin \frac{\theta }{2^{n-2}}}-\frac{1}{2^{n-1}\cdot \sin \frac{\theta }{2^{n-1}}}\right)\\ ={lim}_{n\to \infty }\left\{\frac{1}{2^{r-2}\cdot \sin \frac{\theta }{2^{r-2}}}-\frac{1}{2^{r-1}\cdot \sin \frac{\theta }{2^{r-1}}}\right\}\\ ={lim}_{n\to \infty }\left\{\frac{1}{2^{-1}\cdot \sin 2\theta }-\frac{1}{\sin \theta }+\frac{1}{\sin \theta }-\frac{1}{2\sin \frac{\theta }{2}}+\frac{1}{2\sin \frac{\theta }{2}}-\frac{1}{4\sin \frac{\theta }{4}}\right\}\\ ={lim}_{n\to \infty }\left\{\frac{2}{\sin 2\theta }-\frac{1}{2^{n-1}\sin \frac{\theta }{2^{n-1}}}\right\}\\ =\left(\frac{2}{\sin 2\theta }-\frac{1}{\theta }\right)\\ Hence,optionAiscorrectanswer.\end{array}$