Question

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

Answer 1

We have,

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

Squaring both sides

$\Rightarrow a^2{\cos }^2\theta +b^2{\sin }^2\theta -2ab\sin \theta \cos \theta =c^2$

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

$\Rightarrow a^2-a^2{\sin }^2\theta +b^2-b^2{\cos }^2\theta -2ab\sin \theta \cos \theta =c^2$

$\Rightarrow a^2+b^2-c^2=a^2{\sin }^2\theta +b^2{\cos }^2\theta +2ab\cos \theta \sin \theta$

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

$\Rightarrow (a\sin \theta +b\cos \theta )=\pm \sqrt{a^2+b^2-c^2}\to \left(1\right)$

so

$(a\sin \theta +b\cos \theta )-\sqrt{a^2+b^2-c^2}=0$