Q133 / 240 · KCET 2026 · Mathematics
$\tan^{-1}\left(\dfrac{1}{1 + 1 \cdot 2}\right) + \tan^{-1}\left(\dfrac{1}{1 + 2 \cdot 3}\right) + \ldots + \tan^{-1}\left(\dfrac{1}{1 + n(n+1)}\right) = $
- Correct answerA$\tan^{-1}\left(\dfrac{n}{n+2}\right)$
- B$\tan^{-1}\left(\dfrac{n+1}{n}\right)$
- C$\tan^{-1}\left(\dfrac{n}{n+1}\right)$
- D$\tan^{-1}\left(\dfrac{n+2}{n}\right)$
Solution
The general term of the given series is $T_k = \tan^{-1}\left(\dfrac{1}{1 + k(k+1)}\right)$.
This can be written as $T_k = \tan^{-1}\left(\dfrac{(k+1) - k}{1 + (k+1)k}\right)$.
Using the identity $\tan^{-1}\left(\dfrac{x - y}{1 + xy}\right) = \tan^{-1}x - \tan^{-1}y$, we get:
$T_k = \tan^{-1}(k+1) - \tan^{-1}k$
Writing the terms for $k = 1, 2, \ldots, n$:
$T_1 = \tan^{-1}2 - \tan^{-1}1$
$T_2 = \tan^{-1}3 - \tan^{-1}2$
$T_n = \tan^{-1}(n+1) - \tan^{-1}n$
Adding all the terms, the intermediate terms cancel out:
$S = \sum_{k=1}^{n} T_k = \tan^{-1}(n+1) - \tan^{-1}1$
Applying the inverse tangent difference formula again:
$S = \tan^{-1}\left(\dfrac{(n+1) - 1}{1 + (n+1)(1)}\right)$
$S = \tan^{-1}\left(\dfrac{n}{n+2}\right)$
Answer: $\tan^{-1}\left(\dfrac{n}{n+2}\right)$
This can be written as $T_k = \tan^{-1}\left(\dfrac{(k+1) - k}{1 + (k+1)k}\right)$.
Using the identity $\tan^{-1}\left(\dfrac{x - y}{1 + xy}\right) = \tan^{-1}x - \tan^{-1}y$, we get:
$T_k = \tan^{-1}(k+1) - \tan^{-1}k$
Writing the terms for $k = 1, 2, \ldots, n$:
$T_1 = \tan^{-1}2 - \tan^{-1}1$
$T_2 = \tan^{-1}3 - \tan^{-1}2$
$T_n = \tan^{-1}(n+1) - \tan^{-1}n$
Adding all the terms, the intermediate terms cancel out:
$S = \sum_{k=1}^{n} T_k = \tan^{-1}(n+1) - \tan^{-1}1$
Applying the inverse tangent difference formula again:
$S = \tan^{-1}\left(\dfrac{(n+1) - 1}{1 + (n+1)(1)}\right)$
$S = \tan^{-1}\left(\dfrac{n}{n+2}\right)$
Answer: $\tan^{-1}\left(\dfrac{n}{n+2}\right)$
Hence, the correct answer is option: 'A'.