tan2θ-tanθ和tanθ哪个大同上
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/20 07:50:20
tan2θ-tanθ和tanθ哪个大同上tan2θ-tanθ和tanθ哪个大同上tan2θ-tanθ和tanθ哪个大同上解tan2θ-tanθ-tanθ=tan2θ-2tanθ=2tanθ/(1-ta
tan2θ-tanθ和tanθ哪个大同上
tan2θ-tanθ和tanθ哪个大
同上
tan2θ-tanθ和tanθ哪个大同上
解tan2θ-tanθ-tanθ
=tan2θ-2tanθ
=2tanθ/(1-tan^2θ)-2tanθ
=2tanθ[1/(1-tan^2θ)-1]
=2tanθ[（1-(1-tan^2θ))/(1-tan^2θ)]
=2tanθ[tan^2θ/(1-tan^2θ)]
=2tan^3θ/(1-tan^2θ)
知tanθ＞1时,2tan^3θ/(1-tan^2θ)＜0,即tan2θ-tanθ＜tanθ,
当0＜tanθ＜1时,2tan^3θ/(1-tan^2θ)＞0,即tan2θ-tanθ＞tanθ
当-1＜tanθ＜0时,2tan^3θ/(1-tan^2θ)＜0,即tan2θ-tanθ＜tanθ
当tanθ＜-1时,2tan^3θ/(1-tan^2θ)＞0,即tan2θ-tanθ＞tanθ
tan2θ-tanθ和tanθ哪个大同上
化简 tan2θ-tanθ/1+tan2θtanθ
证明：tanθ-(1/tanθ)=-2/tan2θ
求证：tan3θ-tan2θ-tanθ=tanθtan2θtan3θ
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
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)
已知tanθ=-2,求sin2θ,cos2θ和tan2θ的值前天没听课- -
已知（1-tanθ）/（2+tanθ）=1,求证tan2θ=-4tan(θ+π/4) 想得到就不会来问了！乱死了！
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