Comment
Share
Q)

# The value of $\sum\limits_{n=1}^{\infty}\large\frac{\tan\bigg(\Large\frac{\theta}{2^n}\bigg)}{2^{n-1}\cos\bigg(\Large\frac{\theta}{2^{n-1}}\bigg)}$ is equal to

$\begin{array}{1 1}(a)\;\large\frac{2}{\sin 2\theta}-\frac{1}{\theta}&(b)\;\large\frac{2}{\sin 2\theta}+\frac{1}{\theta}\\(c)\;\large\frac{1}{\sin 2\theta}-\frac{1}{\theta}&(d)\;\large\frac{2}{\sin\theta}-\frac{1}{\theta}\end{array}$

Comment
A)
Step 1:
We have $\sum\limits_{n=1}^{\infty}\large\frac{\tan\bigg(\Large\frac{\theta}{2^n}\bigg)}{2^{n-1}\cos\bigg(\Large\frac{\theta}{2^{n-1}}\bigg)}$
$\Rightarrow \sum\limits_{r=1}^n\large\frac{\sin\big(\Large\frac{\theta}{2^r}\big)}{2^{r-1}\cos\big(\Large\frac{\theta}{2^{r-1}}\big)}$
$\Rightarrow \sum\limits_{r=1}^n\large\frac{2\sin^2\big(\Large\frac{\theta}{2^r}\big)}{2^{r-1}(2\sin\big(\Large\frac{\theta}{2^r}\big)\cos\big(\Large\frac{\theta}{2^r}\big))\cos\big(\Large\frac{\theta}{2^{r-1}}\big)}$
$\Rightarrow \sum\limits_{r=1}^n\large\frac{1-\cos\big(\Large\frac{\theta}{2^{r-1}}\big)}{2^{r-1}\sin\big(\Large\frac{\theta}{2^{r-1}}\big)\cos\big(\Large\frac{\theta}{2^{r-1}}\big)}$
$\Rightarrow \sum\limits_{r=1}^n\bigg[\large\frac{1}{2^{r-2}\sin\big(\Large\frac{\theta}{2^{r-2}}\big)}-\large\frac{1}{2^{r-1}\sin\big(\Large\frac{\theta}{2^{r-1}}\big)}\bigg]$
$\Rightarrow \bigg[\large\frac{2}{\sin 2\theta}-\frac{1}{\sin\theta}\bigg]+\bigg[\large\frac{1}{\sin\theta}-\large\frac{1}{2\sin\big(\Large\frac{\theta}{2}\big)}\bigg]+\bigg[\large\frac{1}{2\sin\Large\frac{\theta}{2}}-\frac{1}{2^2\sin\big(\Large\frac{\theta}{2^2}\big)}\bigg]+\bigg[\large\frac{1}{2^{n-2}\sin\big(\Large\frac{\theta}{2^{n-2}}\big)}-\large\frac{1}{2^{n-1}\sin\big(\Large\frac{\theta}{2^{n-1}}\big)}\bigg]$
$\Rightarrow \large\frac{2}{\sin 2\theta}-\frac{1}{2^{n-1}\sin\big(\Large\frac{\theta}{2^n-1}\big)}$
Step 2:
$\therefore \sum\limits_{n=1}^{\infty}\large\frac{\tan\big(\large\frac{\theta}{2^n}\big)}{2^{n-1}\cos\big(\Large\frac{\theta}{2^{n-1}}\big)}$
$\Rightarrow \lim\limits_{n\to \infty}\sum\limits_{r=1}^n\large\frac{\tan \big(\Large\frac{\theta}{2^r}\big)}{2^{r-1}\cos\big(\Large\frac{\theta}{2^{r-1}}\big)}$
$\Rightarrow \lim\limits_{n\to \infty}\bigg[\large\frac{2}{\sin 2\theta}-\frac{1}{2^{n-1}\sin\big(\Large\frac{\theta}{2^{n-1}}\big)}\bigg]$
$\Rightarrow \large\frac{2}{\sin 2\theta}$$-\lim\limits_{n\to \infty}\bigg[\large\frac{\theta/2^{n-1}}{\theta.\sin\big(\Large\frac{\theta}{2^{n-1}}\big)}\bigg]$
$\Rightarrow \large\frac{2}{\sin 2\theta}-\frac{1}{\theta}$
Hence (a) is the correct option.