The value of tan^-1(2) + tan^-1(3) is equal to

The correct answer is Option (3) → \(\displaystyle\frac{3\pi}{4}\)

Since 2 and 3 are both greater than 1, \(\tan^{-1}(2)\) and \(\tan^{-1}(3)\) both lie between \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\). So their sum lies between \(\frac{\pi}{2}\) and \(\pi\).

The tangent addition formula gives the tangent of the sum:

    \( \displaystyle \tan\left(\tan^{-1}2 + \tan^{-1}3\right) = \frac{2 + 3}{1 - 2 \times 3} \)

    \( \displaystyle = \frac{5}{-5} = -1 \)

The angle between \(\frac{\pi}{2}\) and \(\pi\) whose tangent is −1 is

    \( \displaystyle \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)

Tip: \(\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}\) holds only when \(xy \lt 1\). For \(x, y \gt 0\) with \(xy \gt 1\), add \(\pi\) to it; using the formula alone here gives the trap answer \(-\frac{\pi}{4}\).

Note: This topic was not in the CUET 2026 syllabus.