Prove that, Cosec2θ - Sec2θ = a2/b2 - b2/a2

Question

Prove that, Cosec2θ - Sec2θ = a2/b2 - b2/a2

Answer 1

csc2t - sec2t = a^2/b^2 - b^2/a^2

LS = 1/sin2t - 1/cos2t = (cos2t - sin2t)/sin2tcos2t

RS = (a^4 -b^4)/a^2b^2

let b = sin2t, a = cos2t, then

cos2t -sin2t = cos^4(2t) - sin^4(2t) = (cos^2(2t) +sin^2(2t))(cos2t +sin2t)(cos2t - sin2t)

1 = 1(cos2t + sin2t)

cos2t + sin2t = 1

2t = 0

t = 0 is one solution.

the original equation is not an identity which requires being true for all values of t