We have $a \cos \theta+b \sin \theta=m$ and $a \sin \theta-b \cos \theta=n \quad$...(ii) On squaring and adding Eqns. (i) and (ii), we get
$$
\begin{aligned}
&m^2+n^2=(a \cos \theta+b \sin \theta)^2+(a \sin \theta-b \cos \theta)^2 \\
&=a^2 \cos ^2 \theta+b^2 \sin ^2 \theta+2 a b \sin \theta \cdot \cos \theta+a^2 \sin ^2 \theta \\
&+b^2 \cos ^2 \theta-2 a b \sin \theta \cdot \cos \theta \\
&\Rightarrow \quad m^2+n^2=a^2\left(\cos ^2 \theta+\sin ^2 \theta\right)+b^2\left(\sin ^2 \theta+\cos ^2 \theta\right) \\
&\Rightarrow \quad m^2+n^2=a^2+b^2\left[\because \cos ^2 \theta+\sin ^2 \theta=1\right]
\end{aligned}
$$
Hence proved.