Question

If $a\sin \theta +b\cos \theta =c,$ then prove that $a\cos \theta -b\sin \theta =\pm \sqrt{a^2+b^2-c^2}$

Answer 1

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

Squaring both sides

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

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

Let, $a\cos \theta -b\sin \theta =x$

Squaring both sides

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

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

Add equation (1) and (2),

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

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

$(a^2+b^2)[{\sin }^2\theta +{\cos }^2\theta ]=c^2+x^2$

$(a^2+b^2)=c^2+x^2................[\because {\sin }^{2}x+{\cos }^{2}x=1]$

$(a^2+b^2-c^2)=x^2$

Take square root both sides,

$\pm \sqrt{(a^2+b^2-c^2)}=x$

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