Question

If $a\cos \theta -b\sin \theta =d,$ then the value of $\left(a\sin \theta +b\cos \theta \right)\cdot \left(a\cos \theta -b\sin \theta \right).$

Answer 1

Let $a\sin \theta +b\cos \theta =p$

$\to {(a\sin \theta +b\cos \theta )}^2+{(a\cos \theta -b\sin \theta )}^2=p^2+d^2$

$\to a^2+b^2=p^2+d^2$

$\to p=\sqrt{a^2+b^2-d^2}$

$\to (a\sin \theta +b\cos \theta ).(a\cos \theta -b\sin \theta )=d\sqrt{a^2+b^2-d^2}$