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 .