Question

Prove that sinθsin(90-θ)-cosθcos(90-θ)=0

Answer 1

sinθ sin(90-θ) -cosθ cos (90-θ) = 0
sinθ cosθ - cosθ sinθ = 0
sinθ cosθ - sinθ cosθ = 0
sinθ cosθ = sinθ cosθ
Therefore, LHS = RHS
Hence proved

Answer 2

just change sin(90-Q) to cosQ