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
.