Question

${\tan }^{-1}\frac{1}{1+\left(1\right)\left(2\right)}+{\tan }^{-1}\frac{1}{1+\left(2\right)\left(3\right)}+\dots +{\tan }^{-1}\frac{1}{1+(n-1)\left(n\right)}=$

• A. $0$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. ${\tan }^{-1}(\frac{n-1}{n+1})$

Answer 1

The correct option is D ${\tan }^{-1}(\frac{n-1}{n+1})$

${\tan }^{-1}\frac{1}{1+\left(1\right)\left(2\right)}+{\tan }^{-1}\frac{1}{1+\left(2\right)\left(3\right)}+.......+{\tan }^{-1}\frac{1}{1+(n-1)\left(n\right)}$
$={\tan }^{-1}2-{\tan }^{-1}1+{\tan }^{-1}3-{\tan }^{-1}2+........+{\tan }^{-1}\left(n\right)-{\tan }^{-1}(n-1)$
$={\tan }^{-1}n-{\tan }^{-1}1$
$={\tan }^{-1}\left(\frac{n-1}{1+n}\right)$