How do you find the exact value of # sec(arccos ((-3/5))#?

1 Answer
Sep 17, 2015

#sec(arccos(-3/5)) = -5/3#

Explanation:

Remember that #sec(x) = 1/cos(x)#, so

#sec(arccos(-3/5)) = 1/cos(arccos(-3/5)#

By definition, #cos(arccos(x)) = x#, assuming that #-1 <= x <=1#, since the cosine function can only output values within this range.

So #sec(arccos(-3/5)) = 1/cos(arccos(-3/5)) = 1/(-3/5)#

And that fraction outputs to #-5/3#, so

#sec(arccos(-3/5)) = -5/3#