If in a triangle ABC, sinA = cosB then the value of cosC is?

$$\eqalign{ & {\text{In }}\vartriangle {\text{, }}\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C}} = {\text{18}}{0^ \circ }\,....{\text{(i)}} \cr & {\text{sin A}} = {\text{cos B}} \cr & {\text{sin A}} = {\text{sin}}\left( {{{90}^ \circ } - {\text{B}}} \right){\text{ }} \cr & {\text{A}} = {90^ \circ } - {\text{B}} \cr & {\text{A}} + {\text{B}} = {90^ \circ }\,........(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \angle {\text{C}} = {90^ \circ } \cr & {\text{So, cos C }} = \cos {90^ \circ } = 0 \cr} $$