$\begin{align}  & \text{i)  If }\alpha \text{ and }\beta \text{ satisfy the equation } \\ & a\cos \alpha +b\sin \alpha =c---(i) \\ & a\cos \beta +b\sin \beta =c----(ii) \\ & \text{Multiply }(i)\text{ by }\cos \beta \text{ and }(ii)\text{ to eliminate }a \\ & a\cos \alpha \cos \beta +b\sin \alpha \cos \beta =c\cos \beta ---(iii) \\ & a\cos \alpha \cos \beta +b\cos \alpha \sin \beta =c\cos \alpha ---(iv) \\ & \text{Subtract }(iv)\text{ from (iii)} \\ & b\sin \alpha \cos \beta -b\cos \alpha \sin \beta =c\cos \beta -c\cos \alpha  \\ & b\sin (\alpha -\beta )=c(\cos \beta -\cos \alpha ) \\ & 2b\sin \tfrac{\alpha -\beta }{2}\cos \tfrac{\alpha -\beta }{2}=-2c\sin \tfrac{\alpha +\beta }{2}\sin \tfrac{\alpha -\beta }{2} \\ & \frac{b}{c}=\sin \frac{\alpha +\beta }{2}\times \frac{1}{\cos \tfrac{\alpha -\beta }{2}} \\ & \sin \frac{\alpha +\beta }{2}\cdot \sec \frac{\alpha -\beta }{2}=\frac{b}{c} \\ &  \\ & \text{ii)Eliminate }b\text{ from (i) and (ii) by multiplying (i) by sin}\beta \text{ } \\ & \text{and multiply (ii) by sin}\alpha  \\ & a\cos \alpha \sin \beta +b\sin \alpha \sin \beta =c\sin \beta ---(v) \\ & a\cos \beta \sin \alpha +b\sin \beta \sin \alpha =c\sin \alpha ---(vi) \\ & \text{Subtract (vi) from (v) } \\ & a\cos \alpha \sin \beta -a\cos \beta \sin \beta =c\sin \beta -c\sin \alpha  \\ & a\sin (\beta -\alpha )=c\left[ 2\cos \tfrac{\beta +\alpha }{2}\sin \tfrac{\beta -\alpha }{2} \right] \\ & \frac{c}{a}=\frac{\cos \tfrac{\beta -\alpha }{2}}{\cos \tfrac{\beta +\alpha }{2}} \\ & \frac{\tfrac{a}{c}-1}{\tfrac{c}{a}+1}=\frac{\cos \tfrac{\beta -\alpha }{2}-\cos \tfrac{\beta +\alpha }{2}}{\cos \tfrac{\beta +\alpha }{2}+\cos \tfrac{\beta +\alpha }{2}} \\ & \frac{\tfrac{a}{c}-1}{\tfrac{c}{a}+1}=\frac{-2\sin \tfrac{\beta }{2}\sin (-\tfrac{\alpha }{2})}{2\cos \tfrac{\beta }{2}\cos (-\tfrac{\alpha }{2})}=\frac{2\sin \tfrac{\beta }{2}\sin \tfrac{\alpha }{2}}{2\cos \tfrac{\beta }{2}\cos (-\tfrac{\alpha }{2})} \\ & \frac{c-a}{c+a}=\tan \tfrac{\beta }{2}\tan \tfrac{\alpha }{2} \\\end{align}$