Question

高一数学 同角三角函数 证明题

Answer 1

证明欲证tanxsinx/(tanx-sinx)=(tanx+sinx)/tanxsinx
即需证tan^2xsin^2x=(tanx-sinx)(tanx+sinx)
即需证tan^2xsin^2x=tan^2x-sin^2x
即需证tan^2x-tan^2xsin^2x=sin^2x
即需证tan^2(1-sin^2x)=sin^2x
即需证tan^2xcos^2x=sin^2x
即需证sin^2x/cos^2×cos^2x=sin^2x
即需证sin^2x=sin^2x
而sin^2x=sin^2x显然成立
故原等式成立。

Answer 2

左边=[(sinX)^2/cosX]/[(sinX-sinXcosX)/cosX]
=(sinX)^2/[sinX(1-cos)]
=sinX/(1-cosX)
=sinX(1+cosX)/[1-(cos)^2] 分子与分母都乘以(1+cosX)
=[sinX+sinXcosX]/(sinX)^2
=[tanX+sinX]/sinX*tanX 分子与分母都除以cosX
=右边。

Answer 3

(tanx)^2=(sinx)^2 / (cosx)^2 = (sinx)^2 / (1-(sinx)^2)
(tanx)^2 × (1-(sinx)^2) = (sinx)^2
(tanx)^2 = (tanx)^2 × (sinx)^2 + (sinx)^2
(tanx)^2 - (sinx)^2 = (tanx)^2 × (sinx)^2
平方差公式：
(tanx - sinx) (tanx + sinx) = (tanx sinx)^2
故:
(tanx sinx) / (tanx - sinx) = (tanx + sinx) / (tanx sinx)

Answer 4

左右都除以tanx

Answer 5

交叉相成容易证