tanB=-tan(A+C)=-(tanA+tanC)/(1-tanAtanC)所以tanAtanC=3cos(B+C-A)=cos(pi-A-C+c-A)=cos(pi-2A)=-cos2A=1-2cos^2A=1-2/(1+tan^2A)=1-2/(1+9/tan^2C)=(10-sec^2C)/(8+sec^2C)=(10cos^2C-1)/(8cos^2C+1)=(5cos2C+4)/...