Principal value of tan^-1≤ft(-√3) + tan^-1(1) is:

The correct answer is Option (3) → \(-\displaystyle\frac{\pi}{12}\)

The principal value branch of \(\tan^{-1}\) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).

Since \(\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}\) and \(-\frac{\pi}{3}\) lies in this interval,

    \( \displaystyle \tan^{-1}\left(-\sqrt{3}\right) = -\frac{\pi}{3} \)

Since \(\tan\frac{\pi}{4} = 1\),

    \( \displaystyle \tan^{-1}(1) = \frac{\pi}{4} \)

Adding:

    \( \displaystyle -\frac{\pi}{3} + \frac{\pi}{4} = \frac{-4\pi + 3\pi}{12} = -\frac{\pi}{12} \)