How do you simplify the expression #secthetacos^2theta#?

1 Answer

#secthetacos^2theta=1/costhetacos^2theta=costheta, theta !=pi/2+npi#

Explanation:

Remember that:

#sectheta=1/costheta#

So we can say:

#secthetacos^2theta=1/costhetacos^2theta=costheta#

The constraints of the original still apply to the simplified expression!

So where #secthetacos^2theta# is undefined, so is our simplified version, #costheta#. As is pointed out in the comments below, #sectheta# is undefined where #costheta=0#. Within the range of #[0,2pi)#. That occurs at #pi/2 and (3pi)/2#, and is expressed generally as #pi/2+npi#