Question

在三角形ABC中,∠A、∠B、∠C的对边分别为a、b、c,且满足tanB=cos(C-B)/(sinA+sin(C-B)

Answer 1

tanB=cos(C-B)/(sinA+sin(C-B))即

sinB/cosB=cos(C-B)/[sinA+sin(C-B)]

cosB*cos(C-B)=sinB*[sinA+sin(C-B)]

cosB*cos(C-B)-sinB*sin(C-B)=sinB*sinA

cos[B+(C-B)]=sinB*sinA

cosC=sinB*sinA

cos[180°-（B+A)]=sinB*sinA

-cos(B+A)=sinB*sinA

sinA*sinB-cosA*cosB=sinB*sinA

cosA*cosB=0

∠A=90°或

∠B=90°

直角三角形

第2问没看明白