问题描述： 已知:sin^2A/sin^2B+cos^2Acos^2C=1,求证:tan^2Acot^2B=sin^2C 1个回答 分类：数学 2014-09-21 问题解答： 我来补答 因为cos^2C=(1-sin^2A/sin^2B)/cos^2A所以sin^2C=1-(1-sin^2A/sin^2B)/cos^2A=1+sin^2A/(sin^2B*cos^2A)-1/cos^2A=(sin^2B*cos^2A+sin^2A-sin^2B)/(sin^2B*cos^2A)=[(1-cos^2B)*(1-sin^2A)+sin^2A-1+cos^2B]/(sin^2B*cos^2A)=(sin^2A*cos^2B)/(sin^2B*cos^2A)=tan^2Acot^2B所以等式成立 展开全文阅读
