maths~

設ABCD邊長 = 1 , 則DE = sin θ, CE = cos θ
設MEN // BC , M於AB , N於DC
則tan@ = ME / MB;
tan@ = (1 - NE) / (1 - ND).....*
△CDE = (1/2)CD*NE = (1/2)DE*CE
CD*NE = DE*CE
1*NE = sin θ * cos θ ;
又 ND / DE = sinㄥDEN = sin θ
ND / sin θ = sin θ
ND = sin^2 θ
Sub NE and ND to * :
tan@ = (1 - NE) / (1 - ND)
= (1 - sin θ cos θ) / (1 - sin^2 θ)
= (1 - sin θ cos θ) / (cos^2 θ)