Question
In a triangle $$ABC, \sin A - \cos B = \cos C,$$ then what is $$B$$ equal to ?
A.
$$\pi$$
B.
$$\frac{\pi }{3}$$
C.
$$\frac{\pi }{2}$$
D.
$$\frac{\pi }{4}$$
Answer :
$$\frac{\pi }{2}$$
Solution :
$$\eqalign{
& {\text{In a}}\,\Delta ABC,{\text{we have}} \cr
& \sin A - \cos B = \cos C \cr
& \Rightarrow \sin A = \cos B + \cos C \cr
& \Rightarrow 2\sin \frac{A}{2} \cdot \cos \frac{A}{2} \cr
& = 2\cos \left( {\frac{{B + C}}{2}} \right) \cdot \cos \left( {\frac{{B - C}}{2}} \right)\,\,\,\left[ {\because \sin 2A = 2\sin A \cdot \cos A} \right] \cr
& {\text{and}}\,\,\cos B + \cos C = 2\cos \left( {\frac{{B + C}}{2}} \right) \cdot \cos \left( {\frac{{B - C}}{2}} \right) \cr
& \Rightarrow 2\sin \frac{A}{2} \cdot \cos \frac{A}{2} = 2\cos \left( {{{90}^ \circ } - \frac{A}{2}} \right) \cdot \cos \left( {\frac{{B - C}}{2}} \right) \cr
& \left[ {\because A + B + C = {{180}^ \circ } \Rightarrow \left( {\frac{{B + C}}{2}} \right) = {{90}^ \circ } - \frac{A}{2}} \right] \cr
& \Rightarrow 2\sin \frac{A}{2} \cdot \cos \frac{A}{2} = 2\sin \frac{A}{2} \cdot \cos \left( {\frac{{B - C}}{2}} \right)\,\,\left[ {\because \cos \left( {{{90}^ \circ } - \theta } \right) = \sin \theta } \right] \cr
& \Rightarrow \cos \frac{A}{2} = \cos \left( {\frac{{B - C}}{2}} \right) \cr
& \Rightarrow \frac{A}{2} = \frac{{B - C}}{2} \cr
& \Rightarrow A + C = B\,\,\,.....\left( {\text{i}} \right) \cr
& {\text{Also}},A + C = {180^ \circ } - B\,\,\,.....\left( {{\text{ii}}} \right) \cr
& {\text{So}},{180^ \circ } - B = B \cr
& \Rightarrow 2B = {180^ \circ }\,\,\,\therefore B = {90^ \circ } \cr} $$