tanθ·tan2θ+tan2θ·tan3θ+.+tannθ·tan(n+1)θ要完整过程

化简 tan2θ-tanθ/1+tan2θtanθ tanθ·tan2θ+tan2θ·tan3θ+.+tannθ·tan(n+1)θ要完整过程 化简tanθ·tan2θ+tan2θ·tan3θ+.+tann·θ*tan(n+1)θ 证明：tanθ-(1/tanθ)=-2/tan2θ tan2θ-tanθ和tanθ哪个大同上 求证：tan3θ-tan2θ-tanθ=tanθtan2θtan3θ tanθ-tanθ分之1=-tan2θ分之2 tanθ-tanθ分之1=-tan2θ分之2 (1+tanα)/(1-tanα）=2013,则1/cos2θ+tan2θ=? 已知（1-tanθ）/2+tanθ=1,求证tan2θ=-4tan（θ+π/4） 已知（1-tanθ）/（2+tanθ）=1,求证tan2θ=-4tan(θ+π/4) 求证1/sin2θ + 1/tan2θ +1/sinθ =1/tanθ/2 tanθ=a,求[sin(pi/4+θ)/sin(pi/2-θ)]*tan2θ tanθ=a,求[sin(pi/4+θ)/sin(pi/2-θ)]*tan2θ, 求证1/tanθ-1/tan2θ=1/sin2θ 已知tanθ－cotθ=﹣4,求tan2θ的值 (tanαtan2α)/(tan2α-tanα)化简急 化简tan2αtanα/tan2α-tanα