maths~~~~~

設正方ABCD邊長 = 1 :
ㄥADE = (90 - ㄥEDC) = (90 - (90 - θ)) = θ ,
DE = sin θ ,
在△ADE 中 , 由正弦定理 ,
sin@ / DE = sin ㄥAED / AD
sin@ / sin θ = sin (180° - (@+θ)) / 1
sin@ / sin θ = sin(@ + θ)
sin@ / sin θ = sin@ cosθ + cos@ sinθ
sin@ = sinθ sin@ cosθ + cos@ sin^2 θ
tan@ = tan@ sinθ cosθ + sin^2 θ
(tan@) (1 - sinθ cosθ) = sin^2 θ
tan@ = (sin^2 θ ) / (1 - sinθ cosθ)