已知cosa+cosb=2m,2m的绝对值小于等于1
sina+sinb=2n,2n的绝对值小于等于1
求tga*tgb的值
答案：{(m^2+n^2)^2-m^2}/{(m^2+n^2)^2-n^2}

（如果apple同学不知道“和差化积”请看一下书/问一下老师/同学）

和差化积：
cosa+cosb = 2cos((a+b)/2)cos((a-b)/2)
sina+sinb = 2sin((a+b)/2)cos((a-b)/2)

由条件cosa+cosb = 2m, sina+sinb = 2n,

2cos((a+b)/2)cos((a-b)/2) = 2m
2sin((a+b)/2)cos((a-b)/2) = 2n

所以
cos((a+b)/2)cos((a-b)/2) = m
sin((a+b)/2)cos((a-b)/2) = n

两式相除：
tg((a+b)/2) = n/m.
又因为：cos(a+b) = cos((a+b)/2)^2 - sin((a+b)/2)^2 = (1-tg((a+b)/2)^2)/(1+tg((a+b)/2)^2) = (m*m-n*n)/(m*m+n*n)

cosa*cosb - sina*sinb = (m*m - n*n) / (m*m + n*n) (1)

(cosa+cosb)^2 + (sina+sinb)^2 = 4m^2 + 4n^2
2cosa*cosb + 2sina*sinb = 4m^2 + 4n^2 - 2
cosa*cosb + sina*sinb = 2m^2 + 2n^2 - 1 (2)

结合(1)(2)
2sina*sinb = 2m^2 + 2n^2 - (2m^2)/(m^2+n^2)
sina*sinb = m^2 + n^2 - (m^2)/(m^2+n^2)

2cosa*cosb = 2m^2 + 2n^2 - (2n^2)/(m^2+n^2)
cosa*cosb = m^2 + n^2 - (n^2)/(m^2+n^2)

tga*tgb=(sina*sinb)/(cosa*cosb)
=((m^2+n^2)-(m^2)/(m^2+n^2)) / ((m^2+n^2)-(n^2)/(m^2+n^2))
=((m^2+n^2)^2 - m^2)/((m^2+n^2)^2 - n^2)

好题啊！