Question

If $θ={\tan }^{−1}\left(\frac{2}{9}\right)+{\tan }^{−1}\left(\frac{4}{33}\right)+{\tan }^{−1}\left(\frac{8}{129}\right)+{\tan }^{−1}\left(\frac{16}{513}\right)+…$ up to $∞$, then $\cot θ=$

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

Answer 1

Answer: (B)

$θ={\tan }^{−1}\frac{2}{1+4×2}+{\tan }^{−1}\frac{4}{1+8×4}+$
${\tan }^{−1}\frac{8}{1+16×8}+{\tan }^{−1}\frac{16}{1+32×16}+…$
$={\tan }^{−1}\left(\frac{4−2}{1+4×2}\right)+{\tan }^{−1}\left(\frac{8−4}{1+8×4}\right)$
${\tan }^{−1}\left(\frac{16−8}{1+16×8}\right)+{\tan }^{−1}\left(\frac{32−16}{1+32×16}\right)+…$
$=\left({\tan }^{−1}4−{\tan }^{−1}2\right)+\left({\tan }^{−1}8−{\tan }^{−1}4\right)+$
$\left({\tan }^{−1}16−{\tan }^{−1}8\right)+\left({\tan }^{−1}32−{\tan }^{−1}16\right)+…$ to $∞$
$=\frac{π}{2}−{\tan }^{−1}2$
$⇒\cot \left(\frac{π}{2}−{\tan }^{−1}2\right)=\cot \left({\cot }^{−1}2\right)=2$