Question

If $a\cos \theta +b\sin \theta =m\mathrm{and}a\sin \theta -b\cos \theta =n,\mathrm{prove}\mathrm{tha}t,a{}^2+b^2=m^2+n^2$.

Answer 1

Given:
$$\begin{gathered} a\cos \theta +b\sin \theta =m \\ a\sin \theta -b\cos \theta =n \end{gathered}$$
$$\begin{gathered} \mathrm{RHS}=m^2+n^2={\left(a\cos \theta +b\sin \theta \right)}^2+{\left(a\sin \theta -b\cos \theta \right)}^2 \\ =a^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab\cos \theta \sin \theta +a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\cos \theta \sin \theta \\ =a^2\left({\cos }^2\theta +{\sin }^2\theta \right)+b^2\left({\cos }^2\theta +{\sin }^2\theta \right) \\ =a^2+b^2(\because {\cos }^2\theta +{\sin }^2\theta =1) \\ =\mathrm{LHS} \end{gathered}$$