Question

How do you evaluate `cot ((7pi)/6)`?

Answer 1

It is

`cot((pi+6pi)/6)=cot(pi+pi/6)=cot(pi/6)=sqrt3`

Another way is that

`cot(pi+pi/6)=1/(tan(pi+pi/6))`

Use the `tan(a + b) = [tan a + tan b]/[1 - (tan a)(tan b)]`

so you get:

#tan(pi + pi/6) = [tanpi + tanpi/6]/[1 - (tanpi)(tanpi/6)] = [ 0 + (sqrt3)/3] / [1 - (0)((sqrt3)/3)] = [(sqrt3)/3] / [1 - 0] = [(sqrt3)/3] / 1 = [sqrt3] / 3 #

Hence `cot(pi+pi/6)=1/(sqrt3/3)=sqrt3`