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