If tan²α = Cos²θ - Sin²θ, prove that tan²θ = Cos²α - Sin²α?

Answer 1

Using t for theta and a for alpha:

tan^2(t) = Cos^2(a)-Sin^2(a)
or Sin^2(t)/Cos^2(t) = [Cos^2(a)-Sin^2(a)]/ [Cos^2(a)+Sin^2(a)],
because [Cos^2(a)+Sin^2(a)] = 1
or Cos^2(t)/Sin^2(t) = [Cos^2(a)+Sin^2(a)]/ [Cos^2(a)-Sin^2(a)]
Now using componendo-dividendo
[Cos^2(t) + Sin^2(t)]/[Cos^2(t) -Sin^2(t)] = [2*Cos^2(a)]/[2*Sin^2(a)]
or 1/[Cos^2(t) -Sin^2(t)] = Cos^2(a)/Sin^2(a)
or [Cos^2(t) -Sin^2(t)] = Sin^2(a)/Cos^2(a) = tan^2(a) Proved

Answer 2

Using t for theta and a for alpha:

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

(1 + tan^2(t))tan^2(a) = 1 - tan^2(t)

Rearringing:
(1 + tan^2(a))tan^2(t) = 1 - tan^2(a)

tan^2(t) = (1 - tan^2(a)) / (1 + tan^2(a))
= cos(2a)
= cos^2(a) - sin^2(a).