The principal value of \(\sec^{-1}\left(-\sqrt{2}\right)\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{3\pi}{4}\)
The principal value branch of \(\sec^{-1}\) is \([0, \pi] - \left\{\frac{\pi}{2}\right\}\). A negative secant means a negative cosine, so the angle lies in the second quadrant part of this branch.
Since \(\sec\frac{\pi}{4} = \sqrt{2}\),
\( \displaystyle \sec\left(\pi - \frac{\pi}{4}\right) = -\sec\frac{\pi}{4} = -\sqrt{2} \)
Hence
\( \displaystyle \sec^{-1}\left(-\sqrt{2}\right) = \frac{3\pi}{4} \)
Why the other options are incorrect
Option (1) \(-\frac{\pi}{4}\) — Incorrect. Negative angles are outside \([0, \pi]\), and \(\sec\left(-\frac{\pi}{4}\right) = +\sqrt{2}\).
Option (2) \(\frac{\pi}{4}\) — Incorrect. \(\sec\frac{\pi}{4} = +\sqrt{2}\), not \(-\sqrt{2}\).
Option (4) \(\frac{5\pi}{4}\) — Incorrect. \(\sec\frac{5\pi}{4} = -\sqrt{2}\), but \(\frac{5\pi}{4}\) is greater than \(\pi\).