Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Let $S=\sum _{r=1}^n{\tan }^{-1}\left(\frac{1}{r^2+3r+3}\right)$

So, $T_r={\tan }^{-1}\frac{1}{r^2+3r+2+1}$

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

$={\tan }^{-1}\left(r+2\right)-{\tan }^{-1}\left(r+1\right)$

$T_1={\tan }^{-1}3-{\tan }^{-1}2$

$$\begin{gathered} T_2={\tan }^{-1}4-{\tan }^{-1}3 \\ . \\ . \end{gathered}$$
$T_n={\tan }^{-1}\left(n+2\right)-{\tan }^{-1}\left(n+1\right)$

$\Rightarrow \sum _{r=1}^n{\tan }^{-1}\frac{1}{r^2+3r+3}={\tan }^{-1}\left(n+2\right)-{\tan }^{-1}2$

i.e. $6\tan \left(\underset{n\to \infty }{\lim }\sum _{r=1}^n{\tan }^{-1}\frac{1}{r^2+3r+3}\right)=6\tan \left(\underset{n\to \infty }{\lim }\sum _{r=1}^n\left[{\tan }^{-1}\left(n+2\right)-{\tan }^{-1}2\right]\right)$

$=6\tan \left(\frac{\pi }{2}-{\tan }^{-1}2\right)$

$=6\tan \left({\tan }^{-1}\frac{1}{2}\right)=3$