A geometry problem by Aakhyat Singh

Relevant wiki: Half Angle Tangent Substitution

$\begin{aligned} \frac {\cos A}{1-\sin A} & = \frac {1+\cos A + a\sin A}{1+\cos A + b\sin A} & \small \color{#3D99F6} \text{Using half angle tangent substitution} \\ \frac {\frac {1-t^2}{1+t^2}}{1-\frac {2t}{1+t^2}} & = \frac {1+\frac {1-t^2}{1+t^2}+\frac {2at}{1+t^2}}{1+\frac {1-t^2}{1+t^2}+\frac {2bt}{1+t^2}} & \small \color{#3D99F6} \text{and let }t = \tan \frac A2 \\ \frac {1-t^2}{1+t^2 - 2t} & = \frac {1+t^2+1-t^2 + 2at}{1+t^2+1-t^2 + 2bt} \\ \frac {(1-t)(1+t)}{(1-t)^2} & = \frac {2+ 2at}{2 + 2bt} \\ \frac {1+t}{1-t} & = \frac {1+at}{1+bt} \end{aligned}$

$\implies a=1, b=-1, \implies a+b = 1-1 = \boxed{0}$

U will get a=1,b=-1