Question

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

• A. $c^2-a^2-b^2$
• B. $c^2-a^2+b^2$
• C. $a^2-b^2+c^2$
• D. $a^2+b^2-c^2$

Answer 1

Answer: (D)

$a^2+b^2-c^2$

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

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

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

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

$\Rightarrow c^2+k^2=a^2+b^2$

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

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