If tan^-1(-3x)+tan^-1(-2x)=(π)/4 , then the values of x are

The correct answer is Option (3) → \(\displaystyle\frac{-1}{6}\)

Let \(A = \tan^{-1}(-3x)\) and \(B = \tan^{-1}(-2x)\). Then \(\tan A = -3x\), \(\tan B = -2x\), and \(A + B = \frac{\pi}{4}\).

Taking tan of both sides with the formula for tan(A + B):

    \( \displaystyle \frac{\tan A + \tan B}{1 - \tan A\tan B} = \tan\frac{\pi}{4} = 1 \)

    \( \displaystyle \frac{-3x - 2x}{1 - 6x^2} = 1 \)

    \(-5x = 1 - 6x^2\)
    \(6x^2 - 5x - 1 = 0\)
    \((6x + 1)(x - 1) = 0\)
    \(x = 1 \text{ or } x = -\frac{1}{6}\)

Taking tan of both sides can bring in an extra root: tan(A + B) = 1 also holds when \(A + B = -\frac{3\pi}{4}\). So each root must be checked in the original equation.

x = 1: \(A = \tan^{-1}(-3)\) and \(B = \tan^{-1}(-2)\) are both negative, so A + B is negative and cannot be \(\frac{\pi}{4}\). (In fact \(A + B = -\frac{3\pi}{4}\).) x = 1 is rejected.

\(x = -\frac{1}{6}\): \(A = \tan^{-1}\frac{1}{2}\) and \(B = \tan^{-1}\frac{1}{3}\) both lie in \(\left(0, \frac{\pi}{2}\right)\), and

    \( \displaystyle \tan(A+B) = \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 \)

With A + B between 0 and π, this gives \(A + B = \frac{\pi}{4}\). So \(x = -\frac{1}{6}\) is the only solution.

Why the other options are incorrect

• Option (1) \(-1, \displaystyle\frac{-1}{6}\) — Incorrect. x = −1 is not a root of \(6x^2 - 5x - 1 = 0\) (it gives 6 + 5 − 1 = 10); it makes the left side \(\tan^{-1}3 + \tan^{-1}2 = \frac{3\pi}{4}\).

• Option (2) \(-1, \displaystyle\frac{1}{16}\) — Incorrect. Neither value is a root of \(6x^2 - 5x - 1 = 0\).

• Option (4) \(1, \displaystyle\frac{1}{6}\) — Incorrect. x = 1 is the rejected root, and \(x = \frac{1}{6}\) is not a root at all (the root is \(-\frac{1}{6}\)).

Note 1: It seems that there is a mistake in the original NTA question. The paper printed Option (3) as \(1, \frac{-1}{6}\), but x = 1 does not satisfy the equation (for x = 1 the left side is \(-\frac{3\pi}{4}\), not \(\frac{\pi}{4}\)), so no printed option gave exactly the values of x. For the benefit of students and to ensure that the question is correct and has a single valid answer, Option (3) has been changed to \(\frac{-1}{6}\).

Note 2: This topic was not in the CUET 2026 syllabus.