What is #cot^-1(tan(-(4pi)/3))#?

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Answer 1 · 2017-01-21T18:31:43

#5/6pi#.

Recall that, #tan(-theta)=-tantheta#

#:. cot^-1(tan(-4/3pi))=cot^-1(-tan(4/3pi))#

#=cot^-1(-tan(pi+pi/3))=cot^-1(-tan(pi/3))#

#=cot^-1(-sqrt3)#

Now, we use the Defn. of #cot^-1# function :

#cot^-1x=theta, x in RR iff cottheta=x, theta in (0,pi)#

Since, #cot(pi-pi/6)=-cot(pi/6)=-sqrt3,#

# i.e., cot(5/6pi)=-sqrt3," and, "5/6pi in (0,pi)#

#:. cot^-1(-sqrt3)=5/6pi#.