Using C-R: \(u_x=-\sin x\cosh y=v_y\Rightarrow v=\int-\sin x\cosh y\,dy...\) Alternatively, \(u_y=\cos x\sinh y=-v_x\Rightarrow v_x=-\cos x\sinh y\Rightarrow v=\sin x\sinh y+g(y)\). Then \(v_y=\sin x\cosh y=u_x\)... Actually \(v_y=-\sin x\cosh y\) from C-R. Integrating: \(v=-\sin x\sinh y\). Wait: \(u_x=-\sin x\cosh y=v_y\Rightarrow v=\int-\sin x\cosh y\,dy=-\sin x\sinh y\). Hmm, but then checking: \(u_y=\cos x\sinh y\) must equal \(-v_x=\cos x\sinh y\) ✓. So \(v=-\sin x\sinh y\). However common result for this pair is \(v=\sin x\sinh y\) when \(u=\cos x\cosh y\). Recheck: \(f(z)=\cos z=\cos(x+iy)=\cos x\cosh y - i\sin x\sinh y\). So \(v=-\sin x\sinh y\). Answer: option (A).