Question

The value of ${\tan }^{-1}\frac{4}{7}+{\tan }^{-1}\frac{4}{19}+{\tan }^{-1}\frac{4}{39}+{\tan }^{-1}\frac{4}{67}+\dots \dots ..\infty$ equals

• A. ${\tan }^{-1}1+{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}$
• B. ${\tan }^{-1}1+{\cot }^{-1}3$
• C. ${\cot }^{-1}1+{\cot }^{-1}\frac{1}{2}+{\cot }^{-1}\frac{1}{3}$
• D. ${\cot }^{-1}1+{\tan }^{-1}3$

Answer 1

Answer: (B)

$=7+\frac{(n-1)}{2}[24+8(n-2)]=4n^2+3$
$\therefore T_n^{\prime }={\tan }^{-1}\frac{4}{4n^2+3}={\tan }^{-1}\frac{1}{n^2+\frac{3}{4}}={\tan }^{-1}\frac{1}{1+\left(n^2-\frac{1}{4}\right)}$
$={\tan }^{-1}\left[\frac{\left(n+\frac{1}{2}\right)-\left(n-\frac{1}{2}\right)}{1+\left(n+\frac{1}{2}\right)\left(n-\frac{1}{2}\right)}\right]={\tan }^{-1}\left(n+\frac{1}{2}\right)-{\tan }^{-1}\left(n-\frac{1}{2}\right)$
$\text{ Hence }{S}_{\infty }=\sum _{n=1}^{\infty }{T}_n^{\prime }=\frac{\pi }{2}-{\tan }^{-1}\frac{1}{2}={\tan }^{-1}1+{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}-{\tan }^{-1}\frac{1}{2}={\tan }^{-1}1+{\cot }^{-1}3$