How do you find the exact value of #tan^-1 (-1)#?

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Answer 1 · 2017-08-08T18:06:30

#-pi/4 " or " (3pi)/4#

Well, if

#arctan(-1) = theta#

then

#tantheta = -1#

There are two values of #theta# that satisfy this, according to the unit circle:

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#color(blue)(ulbar(|stackrel(" ")(" "arctan(-1) = -pi/4 " or " (3pi)/4" ")|)#

Answer 2 · 2017-08-08T18:09:50

#tan^-1x=theta, x in RR iff tantheta=x, theta in (-pi/2,pi/2).#
Now, #tan(-pi/4)=-tan(pi/4)=-1, -pi/4 in (-pi/2,pi/2).#

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