How do you simplify the expression #sec(arccos (-3/5))#?

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Answer 1 · 2016-07-04T14:13:16

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

Let #arccos(-3/5)=theta#

Therefore #costheta=-3/5#

and hence #sec(arccos(-3/5))=sectheta#

= #1/costheta=1/(-3/5)#

= #1xx-5/3=-5/3#

Answer 2 · 2016-07-04T16:28:05

#-5/3#

Use cos arc cos x = x.

Here, #sec arc cos (-3/5)=1/(cos arc cos (-3/5))=1/(-3/5)=-5/3#