How do you simplify the expression #sec^2theta-tan^2theta+cot^2theta#?

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Answer 1 · 2016-09-21T13:28:49

This expression simplifies to #csc^2theta#.

Apply the following identities:

•#sectheta = 1/costheta#
•#tantheta = sintheta/costheta#
•#cottheta = costheta/sintheta#

#=1/cos^2theta - sin^2theta/cos^2theta + cos^2theta/sin^2theta#

#=(1 - sin^2theta)/(cos^2theta) + cos^2theta/sin^2theta#

Apply the pythagorean identity #sin^2x + cos^2x = 1 -> 1 - sin^2x = cos^2x#

#=cos^2theta/cos^2theta + cos^2theta/sin^2theta#

#= 1 + cos^2theta/sin^2theta#

#=(sin^2theta + cos^2theta)/sin^2theta#

Apply the identity #sin^2beta + cos^2beta = 1#:

#=1/sin^2theta#

Apply the identity #1/sinalpha = cscalpha#.

#=csc^2theta#

Hopefully this helps!