三角恆等式問題，我計極都計唔到！= =

R.H.S.= (sinΘ + tanΘ)(cosΘ + 1/tanΘ)= sinΘ cosΘ + tanΘ cosΘ + sinΘ(1/tanΘ) + tanΘ(1/tanΘ)= sinΘ cosΘ + sinΘ + cosΘ + 1= sinΘ (cosΘ + 1) + cosΘ + 1= (sinΘ + 1) (cosΘ + 1)= L.H.S.;tan(90°-Θ)/ [1 + tan²(90°-Θ)] = (1/tanΘ) / (1 + 1/tan²Θ)= (1/tanΘ) / [(tan²Θ + 1) / tan²Θ]= tanΘ / (tan²Θ + 1)= (sinΘ/cosΘ) / (sin²Θ/cos²Θ + 1)= (sinΘ/cosΘ) / [(sin²Θ + cos²Θ) / cos²Θ]= (sinΘ/cosΘ) / (1 / cos²Θ)= sinΘ cosΘ