Answer
$${\sin ^2}\theta - {\cos ^2}\theta $$
Work Step by Step
$$\eqalign{
& x = u\sin \theta + v\cos \theta ,{\text{ }}y = u\cos \theta + v\sin \theta \cr
& {\text{Find the partial derivatives }}\frac{{\partial x}}{{\partial u}}{\text{ and }}\frac{{\partial x}}{{\partial v}} \cr
& \frac{{\partial x}}{{\partial u}} = \frac{\partial }{{\partial u}}\left[ {u\sin \theta + v\cos \theta } \right] = \sin \theta \cr
& \frac{{\partial x}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {u\sin \theta + v\cos \theta } \right] = \cos \theta \cr
& {\text{Find the partial derivatives }}\frac{{\partial y}}{{\partial u}}{\text{ and }}\frac{{\partial y}}{{\partial v}} \cr
& \frac{{\partial y}}{{\partial u}} = \frac{\partial }{{\partial u}}\left[ {u\cos \theta + v\sin \theta } \right] = \cos \theta \cr
& \frac{{\partial y}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {u\cos \theta + v\sin \theta } \right] = \sin \theta \cr
& {\text{Calculate the Jacobian}} \cr
& \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = \frac{{\partial x}}{{\partial u}}\frac{{\partial y}}{{\partial v}} - \frac{{\partial y}}{{\partial u}}\frac{{\partial x}}{{\partial v}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting the partial derivatives into }}\left( {\bf{1}} \right) \cr
& \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = \left( {\sin \theta } \right)\left( {\sin \theta } \right) - \left( {\cos \theta } \right)\left( {\cos \theta } \right) \cr
& \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = {\sin ^2}\theta - {\cos ^2}\theta \cr} $$
