We have ABCD cyclic Convex quadrilateral HOW WE CAN PROVE THAT sinABD=sinADC?
The assertion is not true.
I don't think we CAN prove that sinABD = sinADC. In fact, we can show that it is not necessarily true using some counter-examples.
Consider the three convex quadrilaterals inscribed in the same size circles below.
In all three cases angle ABD is the same, but angle ADC is different. How could their sines all be the same if this is the case?
Perhaps you don't mean "cyclic" quadrilateral.