Ta có $PT\Leftrightarrow \frac{sinB+sinC}{sin2B+sin2C}=\frac{sinC}{sin2C}$
$\Leftrightarrow \frac{sinA+sinB+sinC}{sin2A+sin2B+sin2C}=\frac{sinC}{sin2C}=\frac{1}{2cosC} (1)$
Mặt khác ta có
$sinA+sinB+sinC=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}$
$sin2A+sin2B+sin2C=4sinAsinBsinC$
$\Rightarrow (1)\Leftrightarrow \frac{cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}}{sinAsinBsinC}=\frac{1}{2cosC}$
$\Rightarrow cosC=4sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}$
Mà ta lại có $cosA+cosB+cosC=1+4sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}$
Vậy $cosA+cosB=1$