Question

${\tan }^{-1}\left(\frac{1}{1+2}\right)+{\tan }^{-1}\left(\frac{1}{1+\left(2\right)\left(3\right)}\right)+$ ${\tan }^{-1}\left(\frac{1}{1+\left(3\right)\left(4\right)}\right)+....+{\tan }^{-1}\left(\frac{1}{1+n(n+1)}\right)={\tan }^{-1}\theta$, then $\theta =$

• A. $\frac{n}{n+1}$
• B. $\frac{n+1}{n+2}$
• C. $\frac{n+2}{n+1}$
• D. $\frac{n}{n+2}$

Answer 1

The correct option is D $\frac{n}{n+2}$

Given that
${\tan }^{-1}\left(\frac{1}{1+2}\right)+{\tan }^{-1}\left(\frac{1}{1+2.3}\right)+.....+{\tan }^{-1}\left(\frac{1}{1+n(n+1)}\right)={\tan }^{-1}\theta$
${\tan }^{-1}\left(\frac{2-1}{1+\left(2\right)\left(1\right)}\right)+{\tan }^{-1}\left(\frac{3-2}{1+3.2}\right)+....+{\tan }^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)={\tan }^{-1}\theta$
$({\tan }^{-1}2-{\tan }^{-1}1)+({\tan }^{-1}3-{\tan }^{-1}2)+...+\left({\tan }^{-1}(n+1)-{\tan }^{-1}n\right)={\tan }^{-1}\theta$
$\Rightarrow {\tan }^{-1}(n+1)-{\tan }^{-1}1={\tan }^{-1}\theta$
$\Rightarrow {\tan }^{-1}\left(\frac{(n+1)-1}{1+(n+1)}\right)={\tan }^{-1}\theta$
$\Rightarrow \frac{n}{n+2}=\theta$