Evaluate each of the following:
The value of sec is –√2.
∴ The question becomes sec–1(–√2).
Let sec–1(–√2) = y
⇒ sec y = –√2
= – sec = √2
= sec
= sec
The range of principal value of sec–1is [0, π]–{}
and sec = –√2
∴ The principal value of sec–1(–√2) is .