中3一條三角比ga問題...非常急

sinθ X cosθ X tan(90°-θ)+cos^2 (90°-θ)
= sinθ cosθ (1/tanθ) + sin^2 θ
= sinθ cosθ (cosθ / sinθ) + sin^2 θ
= cos^2 θ + sin^2 θ
= 1

sinθ X cosθ Xcosθ+sinθ
=sinθ/cosθXsinθ/cosθ
=1