anyone know this one? i have to make (sin^3+cos^3)/ (1-2cos^2) into (sec-sin)/tan-1?

Question

they are equal to eachother i just have no idea how to even start this. i know all the trig identities but im just baffled

Answer 1

LHS
= (sin³(x) + cos³(x)) / (1 - 2cos²(x))
= (sin(x) + cos(x)(sin²(x) - sin(x)cos(x) + cos²(x)) / (sin²(x) - cos²(x))
= (sin(x) + cos(x))(1 - sin(x)cos(x)) / [(sin(x) + cos(x))(sin(x) - cos(x))]
= (1 - sin(x)cos(x)) / (sin(x) - cos(x))
Divide the numerator and the denominator by cos(x)
(1 - sin(x)cos(x)) / (sin(x) - cos(x))
= (sec(x) - sin(x)) / (1 - tan(x))

Answer 2

sin³(x) + cos³(x) is a sum of cubes. It factors to
[sin(x) + cos(x)][sin²(x) - sin(x)cos(x) + cos²(x)]

Can you further simplify the term in brackets?

Next, replace the 1 in the denominator by sin²(x) + cos²(x) and simplify; you'll be left with a difference of squares, which factors. You should be able to do some cancellation, and the resulting fraction should start looking like your target.