How do you evaluate #arcsec (sec(-pi/3))#?

2 Answers
bp
Apr 19, 2015

#pi/3#

First #sec(-pi/3)# would be equal to #sec(pi/3)# = 2. ( Angle -#pi/3# lies in the fourth quadrant and because cosine is positive there sec would also be positive).

Now the problem is reduced to evaluating arcsec2. Since #sec(pi/3)# is 2, hence arcsec2 would be equal to #pi/3#

Tiago Hands
Apr 19, 2015

#theta=arcsec(sec(-pi/3))#

#sectheta=sec(-pi/3)#

#:.theta=-pi/3#

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