Let us assume 2sin = θ

We know sin

∴ 2sin = 2

⇒ 2sin = √2

∴ The question becomes sec^–1(√2)

Now,

Let sec^–1(√2) = y

⇒ sec y = √2

⇒ sec = √2

The range of principal value of sec^–1is [0, π ]–{}

And sec = √2

∴ The principal value of sec^–1(2sin) is .