(c) Given that $a\,\cos \theta + b\,\sin \theta = m$ 

and $a\,\sin \,\theta – b\,\cos \theta = n.$ 

Squaring and adding, we get

${(a\,\,\cos \theta + b\,\sin \theta )^2} + {(a\,\sin \theta – b\,\cos \theta )^2} = {m^2} + {n^2}$

$ \Rightarrow \,\,{a^2}({\cos ^2}\theta + {\sin ^2}\theta ) + {b^2}({\cos ^2}\theta + {\sin ^2}\theta )$ 

$ + 2ab\,(\cos \theta \,\sin \theta – \sin \theta \,\cos \theta ) = {m^2} + {n^2}$ Hence, ${a^2} + {b^2} = {m^2} + {n^2}.$

Trick : Here we can guess that the value of ${a^2} + {b^2}$ is independent of $\theta$, so put any suitable value of $\theta$

$i.e.$ $\frac{\pi }{2},$ so that $b = m$ and $a = n.$ 

Hence ${a^2} + {b^2} = {m^2} + {n^2}.$

(Also check for other value of $\theta$).