How do you find the exact value of #tan^-1(cos(pi))#?
1 Answer
Aug 2, 2016
Explanation:
Begin with the 'inside function'. That is
#cos(pi)# now
#cos(pi)=-1#
#rArrtan^-1(-1)=-pi/4#
Begin with the 'inside function'. That is
#cos(pi)# now
#cos(pi)=-1#
#rArrtan^-1(-1)=-pi/4#