证明:tanαtan2α+tan2αtan3α+……+tan(n-1)αtan(nα)=tan(nα)/tanα-n
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证明:tanαtan2α+tan2αtan3α+……+tan(n-1)αtan(nα)=tan(nα)/tanα-n
证明:tanαtan2α+tan2αtan3α+……+tan(n-1)αtan(nα)=tan(nα)/tanα-n
证明:tanαtan2α+tan2αtan3α+……+tan(n-1)αtan(nα)=tan(nα)/tanα-n
tan(2α-α)=(tan2α-tanα)/(1+tanαtan2α),tanαtan2α=(tan2α-tanα)/tanα-1
tan2αtan3α=(tan3α-tan2α)/tanα-1┄┈┈tan(n-1)αtan(nα)=(tan(nα)-tan(n-1)α)/tanα-1
tanαtan2α+tan2αtan3α+……+tan(n-1)αtan(nα)=(tan2α-tanα)/tanα-1+(tan3α-tan2α)/tanα-1+┄┄┄+(tan(nα)-tan(n-1)α)/tanα-1=(tan(nα)-tanα)/tanα-(n-1)=tan(nα)/tanα-1+n-1=tan(nα)/tanα-n
tanαtanβ+1=(tanα-tanβ)/tan(α-β)
tanαtan2α+1=(tan2α-tanα)/tan(2α-α)=tan2α/tanα-1
tan2αtan3α+1=(tan3α-tan2α)/tan(3α-2α)
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tan(n-1)αtan(nα)=(tannα-tan(n-1)α)/tan(nα-nα+α)
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tanαtanβ+1=(tanα-tanβ)/tan(α-β)
tanαtan2α+1=(tan2α-tanα)/tan(2α-α)=tan2α/tanα-1
tan2αtan3α+1=(tan3α-tan2α)/tan(3α-2α)
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tan(n-1)αtan(nα)=(tannα-tan(n-1)α)/tan(nα-nα+α)
将上式左右相加
tanαtan2α+tan2αtan3α+……+tan(n-1)αtan(nα)+n-1=tan2α/tanα-1+tan3α/tanα-tan2α/tanα+...+tan(nα)/tanα-tan(n-1)α/tanα
tanαtan2α+tan2αtan3α+……+tan(n-1)αtan(nα)=tan(nα)/tanα-n
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