已知Asec(a)-Ctan(a)=d,Bsec(a)+Dtan(a)=c求证：(A^2)+(B^2)=(C^2)+(D^2)

Question

已知Asec(a)-Ctan(a)=d,Bsec(a)+Dtan(a)=c 求证：(A^2)+(B^2)=(C^2)+(D^2)

Answer 1

令 x=sec(a), y=tan(a).
则 Ax -Cy = D, ①
Bx +Dy = C, ②
x^2 -y^2 = 1. ③

由 ① ② 解得
x = (C^2 +D^2) / (AD +BC),
y = (AC -BD) / (AD +BC).
其中 AD +BC ≠ 0.
代入 ③ 得
(C^2 +D^2)^2 -(AC -BD)^2 = (AD +BC) ^2.
即 0 = (C^2 +D^2) ^2 -[ (AC -BD)^2 +(AD +BC)^2 ]
= (C^2 +D^2)^2 -[ (A^2) (C^2) +(B^2) (D^2) +(A^2) (D^2) +(B^2) (C^2) ]
= (C^2 +D^2) ^2 - (A^2 +B^2) (C^2 +D^2)
= (C^2 +D^2) [ (C^2 +D^2) -(A^2 +B^2) ].
所以 C^2 +D^2 =0,
或 (A^2) +(B^2) = (C^2) +(D^2).

若 C^2 +D^2 =0,
则 C =D =0,
AD +BC =0,
与 AD +BC ≠ 0 矛盾.

综上, (A^2) +(B^2) = (C^2) +(D^2).
= = = = = = = = =
换元法+暴力破解.
暂时只想到这办法,计算可能有误.