证1-tan2x/1+tan^2x=cos^2x-sin^2x

证1-tan2x/1+tan^2x=cos^2x-sin^2x 设n>=2,当x不等于k派,求证：tanx*tan2x+tan2x*tan3x+.+tan(n-1)*tannx=tann 求证：(sin x+cos x)(1-tan x)=(2sin x)/(tan2x) f(x)=tan2x/1-tan²2x 的周期 已知tan(x+y)=2/5,tan(x-y)=-1/4,则tan2x的值是 求证：1/sin2x+1/tan2x+1/sinx=1/tan（x/2） 三角函数：函数y=(1-tan^2x)/(1+tan2x)的最小正周期=? 求证tan(x+y)*tan(x-y)=(tan2x-tan2y)/(1-tan2xtan2y) (1-2sin2xcos2x)/(cos^2 2x-sin^2 2x)=(1-tan2x)/(1+tan2x) 求证：（1-2sin2xcos2x）/(cos^2x-sin^2x)=(1-tan2x)/(1+tan2x) 求证 (1-2sinxcosx)/(cos^2x-sin^2x)=(1-tan2x)/(1+tan2x) 函数y=tan2x/1-tan^22x的最小周期 tan2x=?tan(x+y)=? tan(x-y)=2,tan(x+y)=3 求tan2x/tan2y 若tan(3x+y)=2,tan(x+y)=-4 求tan2x 求tan函数的定义域(1)y=1/1+tanx (2)y=根号tan2x 函数y=cos2x+sin2x/cos2x-sin2x的最小正周期原式为y=1+tan2x/1-tan2x我算到了这里,下面应该是tan（π/4+2x),怎么得到这步的? 1 + tan^2x =?Complete the identity (equation) 1 + tan2x = ?x .Is your answer true for all values of x ?