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 .


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