The value of cos^-1≤ft(cos(6π)/5) + sin^-1≤ft(sin(6π)/5) is

The correct answer is Option (3) → \(\displaystyle\frac{3\pi}{5}\)

\(\frac{6\pi}{5}\) lies outside both principal branches, so each angle is first rewritten as an angle inside the right branch with the same cosine or sine.

For \(\cos^{-1}\), the branch is \([0, \pi]\). Cosine does not change when the angle is replaced by \(2\pi\) minus the angle:

    \( \displaystyle \cos\frac{6\pi}{5} = \cos\left(2\pi - \frac{6\pi}{5}\right) = \cos\frac{4\pi}{5} \)

Since \(\frac{4\pi}{5}\) lies in \([0, \pi]\),

    \( \displaystyle \cos^{-1}\left(\cos\frac{6\pi}{5}\right) = \frac{4\pi}{5} \)

For \(\sin^{-1}\), the branch is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Sine does not change when the angle is replaced by \(\pi\) minus the angle:

    \( \displaystyle \sin\frac{6\pi}{5} = \sin\left(\pi - \frac{6\pi}{5}\right) = \sin\left(-\frac{\pi}{5}\right) \)

Since \(-\frac{\pi}{5}\) lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\),

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

Adding:

    \( \displaystyle \frac{4\pi}{5} - \frac{\pi}{5} = \frac{3\pi}{5} \)