Question

If $a\cos \theta +b\sin \theta =m$ and $a\sin \theta -b\cos \theta =n$ then, $a^2+b^2$ equals

• A. $m+n$
• B. $mn$
• C. $\sqrt{mn}$
• D. $m^2+n^2$

Answer 1

Answer: (D)

$m^2+n^2$

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