(1-tan^2(45°-A))/(1+tan^2(45°-A))=sin2A

Let B = (45-A), substitute sin(B)/cos(B) for tan(B), and the left hand side simplifies to:


\frac {\cos^2B - \sin^2B}{cos^2B + sin^2B}


The numerator and denominator are now both well known trig identities - can you take it from here? And recall that \cos(90-\theta) = sin\theta