Question

$\underset{n\to \infty }{\lim }\left(\tan \theta +\frac{1}{2}\tan \frac{\theta }{2}+\frac{1}{2^2}\tan \frac{\theta }{2^2}+\dots +\frac{1}{2^n}\tan \frac{\theta }{2^n}\right)=$

• A. $\frac{1}{\theta }$
• B. $\frac{1}{\theta }-2\cot 2\theta$
• C. $2\cot 2\theta$
• D. None of these

Answer 1

Answer: (B)

$\tan 2\theta =\frac{\sin 2\theta }{\cos 2\theta }$
$\frac{1}{\cot 2\theta }=\frac{2\sin \theta \cos \theta }{{\cos }^2\theta -{\sin }^2\theta }$
$\cot 2\theta =\frac{{\cos }^2\theta -{\sin }^2\theta }{2\sin \theta \cos \theta }$
$\Rightarrow 2\cot 2\theta =\frac{{\cos }^2\theta }{\sin \theta \cos \theta }-\frac{{\sin }^2\theta }{\sin \theta \cos \theta }$
$\Rightarrow 2\cot 2\theta =\cot \theta -\tan \theta$
$\Rightarrow \tan \theta =\cot \theta -2\cot 2\theta$
Now, $\tan \theta =\cot \theta -2\cot 2\theta$
$\Rightarrow \frac{1}{2}\tan \frac{\theta }{2}=\frac{1}{2}\cot \frac{\theta }{2}-\cot \theta$
$\Rightarrow \frac{1}{2^2}\tan \frac{\theta }{2^2}=\frac{1}{2^2}\cot \frac{\theta }{2}-\frac{1}{2}\cot \theta$
$\Rightarrow \frac{1}{2^n}\tan \frac{\theta }{2^n}=\frac{1}{2^n}\cot \frac{\theta }{2^n}-\frac{1}{2^{n-1}}\cot \frac{\theta }{2^{n-1}}$
$\Rightarrow S=-2\cot 2\theta +\frac{1}{2^n}\cot \frac{\theta }{2^n}$
Therefore,
$\underset{n\to \infty }{\lim }\left(\tan \theta +\frac{1}{2}\tan \frac{\theta }{2}+\frac{1}{2^2}\tan \frac{\theta }{2^2}+\dots +\frac{1}{2^n}\tan \frac{\theta }{2^n}\right)$
$=\underset{n\to \infty }{\lim }S=\underset{n\to \infty }{\lim }\left(-2\cot 2\theta +\frac{1}{2^n}\cot \frac{\theta }{2^n}\right)$
$=-2\cot 2\theta +\underset{n\to \infty }{\lim }\frac{1}{\theta }\left(\frac{\frac{\theta }{2^n}}{\tan \frac{\theta }{2^n}}\right)$
$=-2\cot 2\theta +\frac{1}{\theta }$