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

Answer: Option B

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