tanA+tanB=2(log8 72 + log9 72)
=2(log8 9+log8 8 + log9 8+log9 9)
=2[(2/3)log2 3 +(3/2)log3 2 +2]
所以(sinAcosB+cosAsinB)/cosAcosB=sin(A+B)/cosAcosB
=2[(2/3)log2 3 +(3/2)log3 2 +2]------------①
tanAtanB=-log8 72·log9 72
=-[(2/3)log2 3+1][(3/2)log3 2+1]
=-[(2/3)log2 3 +(3/2)log3 2 +2]
所以sinAsinB/cosAcosB
=-[(2/3)log2 3 +(3/2)log3 2 +2]---------②
由①式和②式得(sinAcosB+cosAsinB)/cosAcosB=-2sinAsinB/cosAcosB
所以sinAcosB+cosAsinB+2sinAsinB=0