If  $a\cos \theta +b\sin \theta =c$, then ${(a\sin \theta -b\cos \theta )}^2=$

Let $(a\sin \theta -b\cos \theta )=x$

$a\cos \theta +b\sin \theta =c$

Squaring on both sides

$a^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab\sin \theta \cos \theta =c^2$…………….(1)

$(a\sin \theta -b\cos \theta )=x$

Squaring on both sides

$a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\sin \theta \cos \theta =x^2$……………………(2)

Add equation (1) and  equation (2)

$a^2+b^2=c^2+x^2$

$\Rightarrow x^2=a^2+b^2-c^2$