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Evaluate each of the following:
cot(–θ) is –cot(θ)
∴ The equation given above becomes cot–1(–cot)
cot.
Therefore
Let cot–1() = y
⇒ cot y =
= cot
The range of principal value of cot–1is (0, π)
and cot
∴ The value of cot–1(cot) is
.