Question

If $a\cos \theta +b\sin \theta =p$, $a\sin \theta -b\cos \theta =q$ prove that $a^2+b^2=p^2+q^2$.

Answer 1

Square both sides of both the given equations and then add,

$a^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab\cos \theta \sin \theta +a^2{\cos }^2\theta +b^2{\sin }^2\theta -2ab\cos \theta \sin \theta =p^2+q^2$

$a^2\left({\cos }^2\theta +{\sin }^2\theta \right)+b^2\left({\cos }^2\theta +{\sin }^2\theta \right)=p^2+q^2$

$a^2+b^2=p^2+q^2$