两道三角恒等变换数学题1.已知α、β均为锐角,且cos(α+β)=sin(α-β),则tanα=_____.2.(1+tan21°)(1+tan22°)(1+tan23°)(1+tan24°)的值是_____.
问题描述：
两道三角恒等变换数学题
1.已知α、β均为锐角,且cos(α+β)=sin(α-β),则tanα=_____.
2.(1+tan21°)(1+tan22°)(1+tan23°)(1+tan24°)的值是_____.
答
1.由于：cos(A+B)=sin(A-B)则有：[cosAcosB-sinAsinB]=[sinAcosB-sinBcosA][cosBcosA-cosBsinA]=[sinBsinA-sinBcosA]-cosB(sinA-cosA)+sinB(sinA-cosA)=0(sinA-cosA)(sinB-cosB)=0则：sinA-cosA=0sinA=cosAtanA=12....