Generalize this (part-2)

Geometry Level 3

$\large\tan^{-1} \left(\dfrac{1}{1 + 1\cdot2}\right) + \tan^{-1} \left(\dfrac{1}{1 + 2\cdot3}\right) + \cdots +\tan^{-1} \left(\dfrac{1}{1 + n(n+1)}\right)$ If above expression is equal to $\tan^{-1}\theta$ ,then find $\theta$ .

\dfrac{\pi}{4}

\dfrac{n}{n+2}

\dfrac{\pi}{2}

None of these

\dfrac{n+1}{n+2}

1 solution

Akhil Bansal
Dec 11, 2015

$\large = \displaystyle \sum_{n=1}^{n} \tan^{-1}\left(\dfrac{1}{1 + n(n+1)}\right)$ $\large = \displaystyle \sum_{n=1}^{n} \tan^{-1}\left(\dfrac{n + 1 - n}{1 + n(n+1)}\right)$

Using trigonometry identity,
$\tan^{-1}\left( \dfrac{a - b }{1 + ab} \right) = \tan^{-1} a - \tan^{-1} b$

$\large = \displaystyle \sum_{n=1}^{n} \left(\tan^{-1}\left(n+1\right) - \tan^{-1}( n)\right)$ $\large = \tan^{-1} \left(n+1\right) - \tan^{-1}(1)$ $\large = \tan^{-1} \left(\dfrac{n}{n+2}\right)$

Moderator note:

Can you state the important inverse tangent property that you repeatedly used here?

Generalize this (part-2)

1 solution

Moderator note:

0 pending reports