for given Triangle
sin A = cos B is always
Costuet
Hence option d is career
[tex]given\ \ \ \ \ < C=90\textdegree[/tex]
[tex]In\ \Delta\ ABC;\ \ \ \ \ < A+ < B+ < C=180[/tex]
[tex]A+B=180-C[/tex]
[tex]A+B=180-90=90[/tex]
[tex]A=90-B[/tex]
[tex]taking\ sine\ on\ noth\ side[/tex]
[tex]sin\ A=sin(90-B)\\ sin\ A=cos\ B[/tex]
[tex]Henee\ ophon\ 'd'\ is\ cgrrect.[/tex]
I hope this helps you
:)