Question

${\tan }^{-1}\left(\frac{2}{4}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)+{\tan }^{-1}\left(\frac{2}{16}\right)+{\tan }^{-1}\left(\frac{2}{25}\right)+\dots \dots \dots \infty$ equals

• A. $\pi -{\tan }^{-1}(1)$
• B. $\frac{\pi }{2}-{\tan }^{-1}(1)$
• C. $\pi -{\tan }^{-1}(3)$
• D. ${\tan }^{-1}(3)$

Answer 1

Answer: (D)

$T_1={\tan }^{-1}3-{\tan }^{-1}1$
$T_2={\tan }^{-1}4-{\tan }^{-1}2$
$T_3={\tan }^{-1}5-{\tan }^{-1}3$
$T_{n-1}={\tan }^{-1}(n+1)-{\tan }^{-1}(n-1)$
$T_n={\tan }^{-1}(n+2)-{\tan }^{-1}(n)$
$\therefore S_{\infty }=\pi -\left({\tan }^{-1}(1)+{\tan }^{-1}(2)\right)={\tan }^{-1}(3)<br/>$