The value of tan^-1≤ft(tan(3π)/4)+2cos^-1≤ft(cos(7π)/6) is:

The correct answer is Option (1) → \(\displaystyle\frac{17\pi}{12}\)

\(\tan^{-1}(\tan\theta) = \theta\) and \(\cos^{-1}(\cos\theta) = \theta\) hold only when \(\theta\) is in the principal value branch: \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) for tan⁻¹ and \([0, \pi]\) for cos⁻¹. Neither \(\frac{3\pi}{4}\) nor \(\frac{7\pi}{6}\) is in its branch, so each angle must first be replaced by one in the branch with the same tan or cos.

For the first term, tan has period \(\pi\):

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

    \(\displaystyle \tan^{-1}\left(\tan\frac{3\pi}{4}\right) = -\frac{\pi}{4}\)

For the second term, \(\cos(2\pi-\theta) = \cos\theta\):

    \(\displaystyle \cos\frac{7\pi}{6} = \cos\left(2\pi-\frac{7\pi}{6}\right) = \cos\frac{5\pi}{6}\)

    \(\displaystyle \cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}\)

Adding:

    \(\displaystyle -\frac{\pi}{4}+2\times\frac{5\pi}{6} = -\frac{3\pi}{12}+\frac{20\pi}{12}\)

    \(\displaystyle = \frac{17\pi}{12}\)

Why the other options are incorrect

• Option (2) \(\displaystyle\frac{37\pi}{12}\) — Incorrect. This is \(\frac{3\pi}{4}+2\times\frac{7\pi}{6}\), cancelling the functions without checking the principal branches.

• Option (3) \(-\displaystyle\frac{25\pi}{12}\) — Incorrect. The value is \(-\frac{\pi}{4}+\frac{5\pi}{3} = \frac{17\pi}{12}\), which is positive.

• Option (4) \(-\displaystyle\frac{31\pi}{12}\) — Incorrect. The value is \(\frac{17\pi}{12}\), which is positive.