How do you evaluate #tan^-1(tan((5pi)/6))#?

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Answer 1 · 2016-08-04T17:38:22

#=5pi/6#

#tan^-1(tan(5pi/6))#
#=5pi/6#

Answer 2 · 2016-08-04T20:00:48

#tan^(-1)(tan((5pi)/6)) = -pi/6#

#theta = tan^(-1)(tan((5pi)/6))# by definition satisfies both of the conditions:

Note that #tan# has period #pi#, so for any integer #n#:

#tan ((5pi)/6 + npi) = tan ((5pi)/6)#

When #n = -1#, we have:

#(5pi)/6+npi = (5pi)/6 - pi = -pi/6#

which lies in the range #(-pi/2, pi/2)#, so satisfies the second condition for #tan^(-1)#

Thus:

#tan^(-1)(tan((5pi)/6)) = -pi/6#