Question

$\displaystyle\lim _{n \rightarrow \infty}\left(\tan \theta+\frac{1}{2} \tan \frac{\theta}{2}+\frac{1}{2^{2}} \tan \frac{\theta}{2^{2}}+\ldots+\frac{1}{2^{n}} \tan \frac{\theta}{2^{n}}\right)=$

• 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} $
$\Rightarrow \cot 2 \theta =\frac{\cos ^{2} \theta-\sin ^{2} \theta}{2 \sin \theta \cos \theta}$
$\Rightarrow 2 \cot 2 \theta=\frac{\cos ^{2} \theta}{\sin \theta \cos \theta}-\frac{\sin ^{2} \theta}{\sin \theta \cos \theta}$
$\Rightarrow 2 \cot 2 \theta=\cot \theta-\tan \theta$
$\Rightarrow \tan \theta=\cot \theta-2 \cot 2 \theta$
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,
$ \displaystyle\lim _{n \rightarrow \infty} \left(\tan \theta+\frac{1}{2} \tan \frac{\theta}{2}+\frac{1}{2^{2}} \tan \frac{\theta}{2^{2}}+\ldots+\frac{1}{2^{n}} \tan \frac{\theta}{2^{n}}\right) $
$=\displaystyle\lim _{n \rightarrow \infty} S=\displaystyle\lim _{n \rightarrow \infty}\left(-2 \cot 2 \theta+\frac{1}{2^{n}} \cot \frac{\theta}{2^{n}}\right) $
$=-2 \cot 2 \theta+\displaystyle\lim _{n \rightarrow \infty} \frac{1}{\theta}\left(\frac{\frac{\theta}{2^{n}}}{\tan \frac{\theta}{2^{n}}}\right) $
$=-2 \cot 2 \theta+\frac{1}{\theta} $