Answer:
Let θ = (45 - A)°.
=> 2Tanθ/ 1 + Tan²θ
=> 2Sinθ/Cosθ / 1 + Sin²θ / Cos²θ
=> 2Sinθ/Cosθ/ Cos²θ + Cos²θ / Cos²θ
=> 2Sinθ/Cosθ/ 1 / Cos²θ
=> 2SinθCos²θ/Cosθ
=> 2SinθCosθ
=> Sin2θ
But we assumed θ = (45 - A)°,
=> Sin2(45 - A)
=> Sin(90 - 2A) (∵ Sin(90 - θ) = Cosθ)
=> Cos2A
= R.H.S
Hence proved.
Step-by-step explanation:
