How do you simplify the expression #(sec^2theta-1)/sec^2theta#?

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Answer 1 · 2016-10-30T21:39:59

#=>(1/cos^2theta - 1)/(1/cos^2theta)#

#=> (1/cos^2theta - cos^2theta/cos^2theta)/(1/cos^2theta)#

#=>((1 - cos^2theta)/cos^2theta)/(1/cos^2theta)#

We use the identity #sin^2beta + cos^2beta = 1-> sin^2beta = 1 - cos^2beta#.

#=>(sin^2theta)/cos^2theta xx cos^2theta/1#

#=>sin^2theta#

Hopefully this helps!