How do you verify #(sin^3 t + cos^3 t)/(cos2t) = (sec^2 t - tant)/(sect - tant*sect)#?

1 Answer
Nov 26, 2015

Apply identities, and use the difference of square and sum of cubes formulas.

Explanation:

We will be using the following:
#cos(2t) = cos^2(t) - sin^2(t)#
#sin^2(t) + cos^2(t) = 1#
#sec(t) = 1/cos(t)#
#tan(t) = sin(t)/cos(t)#
#a^2 - b^2 = (a+b)(a-b)#
#a^3 + b^3 = (a+b)(a^2 -ab + b^2)#


Starting from the right hand side:

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

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

#= (1 - sin(t)cos(t))/(cos(t) - sin(t))#

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

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

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

#=(sin^3(t) + cos^3(t))/(cos^2(t)-sin^2(t))#

#= (sin^3(t) + cos^3(t))/cos(2t)#