Evaluate each of the following:

cot(–θ) is –cot(θ)

∴ The equation given above becomes cot^–1(–cot)

cot = 1.

⇒ –cot = –1.

∴ we get cot^–1(–1)

Let cot^–1(–1) = y

⇒ cot y = –1

= – cot = 1

= cot

= cot

The range of principal value of cot^–1is (0, π)

and cot = –1

∴ The value of cot^–1(cot) is .