sinB=3/5,B为钝角,且sin(A+B)=cosA,则tan(A+B)=

题目

sinB=3/5,B为钝角,且sin(A+B)=cosA,则tan(A+B)=

答案

sin(A+B)=cosA,sin(A+B)=cos[(A+B)-B],sin(A+B)=cos(A+B) cos B+ sin(A+B) sin B,sinB=3/5,B为钝角,则cos B=-4/5,代入上式得：sin(A+B)=cos(A+B)•( -4/5)+ sin(A+B) •(3/5) ,sin(A+B) •(2/5)= cos...