How do you prove #cos2t=(1-tan^2 t)/(1+tan^2 t)#?

1 Answer
Alexander
Jul 19, 2016

Remembering that #sec^2(t)-1=tan^2(t)# and the identity
#cos(2t) = cos^2(t) - sin^2(t)# as well as each of the reciprocal identities, namely

#tan t = sin t / cos t# and #sec t = 1/cos t#, we can start off by simplifying our equation on the right-hand side.

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

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

Related questions
See all questions in Proving Identities
Impact of this question
3090 views around the world
You can reuse this answer
Creative Commons License