已知tan(a-r)/tana+sin^2b/sin^2a=1 ,求证tan^2b=tana*tanr..

分析法倒推
tanr=-tan(a-r-a)=[tana-tan(a-r)]/[1+tana*tan(a-r)]
tana*tanr=[tan^2a-tana*tan(a-r)]/[1+tana*tan(a-r)]
=[1+tan^2a-tana*tan(a-r)-1]/[1+tana*tan(a-r)]
=sec^2a/[1+tana*tan(a-r)]-1 (seca=1/cosa)
要证明所证等式,即要证明 sec^2b=sec^2a/[1+tana*tan(a-r)]
即 1+tana*tan(a-r)=cos^2b/cos^2a
即 tana*tan(a-r)=(cos^2b-cos^2a)/cos^2a
即 tan(a-r)/tana=(sin^2a-sin^2b)/sin^2a
=1-sin^2b/sin^2a
两边不等式整理移项即可得已知等式