If in a $$\Delta \,ABC,\cos A\sin B = \sin C$$      then the value of $$\tan \frac{A}{2},$$  if $$3b - 5c = 0,$$   is

Solution :
$$\eqalign{ & \cos A\sin B = \sin C \cr & \Rightarrow \sin \left( {A + B} \right) - \sin \left( {A - B} \right) = 2\sin C \cr & \Rightarrow \sin C = \sin \left( {B - A} \right) \cr & \Rightarrow A + C = B\,\left( {\because A + B = \pi - C} \right) \cr & \therefore B = \frac{\pi }{2} \cr & {\text{Now, }}3b - 5c = 0 \cr & \Rightarrow 3 - 5\sin C = 0 \cr & \therefore \sin C = \frac{3}{5}\,{\text{and}}\,A = \frac{\pi }{2} - C \cr & \Rightarrow \cos A = \sin C \cr & \Rightarrow \frac{{1 - {{\tan }^2}\frac{A}{2}}}{{1 + {{\tan }^2}\frac{A}{2}}} = \frac{3}{5} \cr & \therefore {\tan ^2}\frac{A}{2} = \frac{1}{4} \cr & \Rightarrow \tan \frac{A}{2} = 0.5 \cr} $$