Correct option is (B) sinA
\(\cos \left(90^{\circ}-A\right)=\sin A\)
\(\cos (A-B)=\cos A \cos B+\sin A \sin B\)
Let \(A=90^{\circ}\) and \(B=a\)
Therefore,
\(\cos \left(90^{\circ}-a\right)=\cos 90^{\circ} \cos a+\sin 90^{\circ} \sin a\)
Here, we have,
\(\cos 90^{\circ} =0 \text { and } \sin 90^{\circ}=1\)
\(\cos \left(90^{\circ}-a\right) =\sin a\)
\(\therefore \cos \left(90^{\circ}-A\right) =\sin A\)
