Prove that sinθsin(90-θ)-cosθcos(90-θ)=0
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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
sinθ cosθ - cosθ sinθ = 0
sinθ cosθ - sinθ cosθ = 0
sinθ cosθ = sinθ cosθ
Therefore, LHS = RHS
Hence proved
Anurag19:
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just change sin(90-Q) to cosQ
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