Q64 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
If $\tan^{-1}(1) + \tan^{-1}(3) + \tan^{-1}(5) + \tan^{-1}\left(\dfrac{1}{4}\right) = \pi + \tan^{-1}\left(\dfrac{\alpha}{2}\right)$, then the value of $\alpha$ is...
- Correct answerA$\dfrac{46}{41}$
- B$\dfrac{23}{41}$
- C$\dfrac{42}{41}$
- D$\dfrac{44}{41}$
Solution
Grouping the terms of the left hand side:
$\tan^{-1}(1) + \tan^{-1}\left(\dfrac{1}{4}\right) = \tan^{-1}\left(\dfrac{1 + \dfrac{1}{4}}{1 - \dfrac{1}{4}}\right) = \tan^{-1}\left(\dfrac{5}{3}\right)$
Adding $\tan^{-1}(3)$ to this result:
$\tan^{-1}\left(\dfrac{5}{3}\right) + \tan^{-1}(3) = \pi + \tan^{-1}\left(\dfrac{\dfrac{5}{3} + 3}{1 - \dfrac{5}{3} \times 3}\right)$
$= \pi + \tan^{-1}\left(\dfrac{14}{-12}\right) = \pi - \tan^{-1}\left(\dfrac{7}{6}\right)$
Adding $\tan^{-1}(5)$ to this result:
$\pi - \tan^{-1}\left(\dfrac{7}{6}\right) + \tan^{-1}(5) = \pi + \tan^{-1}\left(\dfrac{5 - \dfrac{7}{6}}{1 + 5 \times \dfrac{7}{6}}\right)$
$= \pi + \tan^{-1}\left(\dfrac{\dfrac{23}{6}}{\dfrac{41}{6}}\right) = \pi + \tan^{-1}\left(\dfrac{23}{41}\right)$
Equating this to the right hand side:
$\pi + \tan^{-1}\left(\dfrac{23}{41}\right) = \pi + \tan^{-1}\left(\dfrac{\alpha}{2}\right)$
$\dfrac{\alpha}{2} = \dfrac{23}{41} \Rightarrow \alpha = \dfrac{46}{41}$
Answer: $\dfrac{46}{41}$
$\tan^{-1}(1) + \tan^{-1}\left(\dfrac{1}{4}\right) = \tan^{-1}\left(\dfrac{1 + \dfrac{1}{4}}{1 - \dfrac{1}{4}}\right) = \tan^{-1}\left(\dfrac{5}{3}\right)$
Adding $\tan^{-1}(3)$ to this result:
$\tan^{-1}\left(\dfrac{5}{3}\right) + \tan^{-1}(3) = \pi + \tan^{-1}\left(\dfrac{\dfrac{5}{3} + 3}{1 - \dfrac{5}{3} \times 3}\right)$
$= \pi + \tan^{-1}\left(\dfrac{14}{-12}\right) = \pi - \tan^{-1}\left(\dfrac{7}{6}\right)$
Adding $\tan^{-1}(5)$ to this result:
$\pi - \tan^{-1}\left(\dfrac{7}{6}\right) + \tan^{-1}(5) = \pi + \tan^{-1}\left(\dfrac{5 - \dfrac{7}{6}}{1 + 5 \times \dfrac{7}{6}}\right)$
$= \pi + \tan^{-1}\left(\dfrac{\dfrac{23}{6}}{\dfrac{41}{6}}\right) = \pi + \tan^{-1}\left(\dfrac{23}{41}\right)$
Equating this to the right hand side:
$\pi + \tan^{-1}\left(\dfrac{23}{41}\right) = \pi + \tan^{-1}\left(\dfrac{\alpha}{2}\right)$
$\dfrac{\alpha}{2} = \dfrac{23}{41} \Rightarrow \alpha = \dfrac{46}{41}$
Answer: $\dfrac{46}{41}$
Hence, the correct answer is option: 'A'.