Find the Principal Value of sec-1(−√2)
Solution:
Let
\[ y = \sec^{-1}(-\sqrt{2}) \]
Then,
\[ \sec y = -\sqrt{2} \]
Taking reciprocal:
\[ \cos y = -\frac{1}{\sqrt{2}} \]
We know:
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
So,
\[ \cos y = \cos\left(\pi – \frac{\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) \]
Since principal value range of sec-1(x) is:
\[ [0,\pi], \quad y \ne \frac{\pi}{2} \]
Therefore,
\[ y = \frac{3\pi}{4} \]
Final Answer:
Principal Value = \[ \frac{3\pi}{4} \]