How do you simplify the expression #(1-tan^2t)/(1+tan^2t)+1#?

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Answer 1 · 2016-10-11T13:24:10

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

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

#=(cos^2t - sin^2t)/(cos^2t) xx cos^2t/(cos^2t + sin^2t)#

Apply the identity #sin^2theta + cos^2theta = 1#.

#=cos^2t - sin^2t #

#=1 - sin^2t - sin^2t#

#= 1 - 2sin^2t#

Apply the double angle identity #cos2theta = 1 - 2sin^2theta#.

#=cos2t#

Hopefully this helps!