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Question
Evaluate the following:
`cot^-1(cot (19pi)/6)`
Solution
We know that
cot-1 (cot θ) = θ, (0, π)
We have
`cot^-1(cot (19pi)/6)=cot^-1[cot(pi+pi/6)]`
`=cot^-1(cot pi/6)`
`=pi/6`
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