用数学归纳法证明tana·tan2a+tan2a·tan3a+…+tan(n-1)a·tanna=tanna/tana-n(n≥2,n∈N+)

证：(1)当n=2时,因为tan2a=2tana/[1-(tana)^2] ,
故左式=tana tan2a=2(tana)^2/[1-(tana)^2]
=-2+2/[1-(tana)^2]
=-2+2tana/[1-(tana)^2].tana
=-2+tan2a/tana=右式,满足题意
2)假设当n=k-1 ,k≥3时,tana·tan2a+tan2a·tan3a+…+tan(k-2)a·tan(k-1)a=tan(k-1)a/tana-(k-1)
则当n=k时,tana·tan2a+tan2a·tan3a+…+tan(k-1）a·tanka
=tan(k-1)a/tana-(k-1)+tan(k-1）a·tanka
=tan(k-1)a/tana-k+1+tan(k-1）a·tanka
又tana=tan[ka-(k-1)a]
=[tanka -tan(k-1)]/[1+tanka·tan(k-1)a]
因此 1+tanka·tan(k-1)a =[tanka -tan(k-1)]/tana
即tan(k-1)a/tana+1+tan(k-1）a·tanka=tanka/tana
故tana·tan2a+tan2a·tan3a+…+tan(k-1）a·tanka
=tan(k-1)a/tana-k+1+tan(k-1）a·tanka
=tan(k-1)a/tana+1+tan(k-1）a·tanka-k
=tanka-k
故得证