Question

If $a\cos \theta -b\sin \theta =c$, then $a\sin \theta -b\cos \theta$ equal to

• A. $\pm \sqrt{a^2+b^2-c^2}$
• B. $\pm \sqrt{a^2+b^2+c^2}$
• C. $\pm \sqrt{a^2-b^2-c^2}$
• D. None of these

Answer 1

The correct option is A $\pm \sqrt{a^2+b^2-c^2}$

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

$a\sin \theta -6\cos \theta =?$

let us assume an expression.

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

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

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

$x=a^2+b^2$

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

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

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

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