1 求证：tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)

1.
两角和正切公式：
tan[(x-y)+(y-z)]=[tan(x-y)+tan(y-z)]/[1-tan(x-y)tan(y-z)]
tan(x-y)+tan(y-z)
=tan(x-y+y-z)*[1-tan(x-y)tan(y-z)]
=tan(x-z)*[1-tan(x-y)tan(y-z)]
=tan(x-z)-tan(x-z)tan(x-y)tan(y-z)
=-tan(z-x)+tan(z-x)tan(x-y)tan(y-z)
tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)
2.
证明:∵x+y+z=nπ（n∈Z）
∴x+y=nπ-z
∴tan(x+y)=tan(nπ-z)
又∵tan(x+y)=(tan x + tan y)/(1-tanx tany)
tan(nπ-z)=-tan z
∴tan x + tan y =-tanz+ tan x tan y tan z
∴tan x+tan y+tan z=tan x*tan y*tan z

1,
若tan(x-y)tan(y-z) = 1
则，
tan(x-y) = cot(y-z) = tan(PI/2 - y + z)
x - y = kPI + PI/2 - y + z, k为任意整数。
x - z = kPI + PI/2,
tan(x-z)为无穷大。矛盾。
所以，
tan(x-y)tan(y-z) 不等于 1...

您好！您的问题是这样的。用两倍角正切公式的变形，得tan(x-y)+tan(y-z)+tan(z-x)＝tan[(x-y)+(y-z)]*[1-tan(x-y)tan(y-z)]+tan(z-x)＝tan(x-z)*[1-tan(x-y)tan(y-z)]+tan(z-x)＝[tan(x-z)+tan(z-x)]-tan(x-z)tan(x-y)tan(y-z)=0+tan(x-y)tan(y-z...