Question

If $\theta=\tan ^{-1}\left(\frac{2}{9}\right)+\tan ^{-1}\left(\frac{4}{33}\right)+\tan ^{-1}\left(\frac{8}{129}\right)+\tan ^{-1}\left(\frac{16}{513}\right)+\ldots$ up to $\infty$, then $\cot \theta=$

• A. $1$
• B. $2$
• C. $\sqrt{3}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (B)

$\theta=\tan ^{-1} \frac{2}{1+4 \times 2}+\tan ^{-1} \frac{4}{1+8 \times 4}+$
$\tan ^{-1} \frac{8}{1+16 \times 8}+\tan ^{-1} \frac{16}{1+32 \times 16}+\ldots$
$=\tan ^{-1}\left(\frac{4-2}{1+4 \times 2}\right)+\tan ^{-1}\left(\frac{8-4}{1+8 \times 4}\right)$
$\tan ^{-1}\left(\frac{16-8}{1+16 \times 8}\right)+\tan ^{-1}\left(\frac{32-16}{1+32 \times 16}\right)+\ldots$
$=\left(\tan ^{-1} 4-\tan ^{-1} 2\right)+\left(\tan ^{-1} 8-\tan ^{-1} 4\right)+$
$\left(\tan ^{-1} 16-\tan ^{-1} 8\right)+\left(\tan ^{-1} 32-\tan ^{-1} 16\right)+\ldots$ to $\infty$
$=\frac{\pi}{2}-\tan ^{-1} 2$
$\Rightarrow \cot \left(\frac{\pi}{2}-\tan ^{-1} 2\right)=\cot \left(\cot ^{-1} 2\right)=2$