If $$x\cos \theta + y\sin \theta = z,$$     then what is the value of $${\left( {x\sin \theta - y\cos \theta } \right)^2}?$$

Solution :
$$\eqalign{ & {\text{Here, }}z = x\cos \theta + y\sin \theta \cr & {z^2} = {x^2}{\cos ^2}\theta + {y^2}{\sin ^2}\theta + 2\,xy\sin \theta \cos \theta \cr & \Rightarrow 2\,xy\sin \theta \cos \theta = {z^2} - {x^2}{\cos ^2}\theta - {y^2}{\sin ^2}\theta \cr & {\text{Let, }}L = {\left( {x\sin \theta - y\cos \theta } \right)^2} \cr & \Rightarrow L = {x^2}{\sin ^2}\theta + {y^2}{\cos ^2}\theta - 2\,xy\sin \theta \cos \theta \cr & \Rightarrow L = {x^2}{\sin ^2}\theta + {y^2}{\cos ^2}\theta - \left[ {{z^2} - {x^2}{{\cos }^2}\theta - {y^2}{{\sin }^2}\theta } \right] \cr & \Rightarrow L = {x^2}\left[ {{{\sin }^2}\theta + {{\cos }^2}\theta } \right] + {y^2}\left[ {{{\sin }^2}\theta + {{\cos }^2}\theta } \right] - {z^2} \cr & \Rightarrow L = {x^2} + {y^2} - {z^2} \cr} $$