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

The correct option is D $m^2+n^2$

Squaring $m$ and $n$ and adding them, we get
$m^2+n^2=(acos\theta +bsin\theta {)}^2+(asin\theta -bcos\theta {)}^2$
$=a^{2}co{s}^2\theta +b^{2}si{n}^2\theta +2absin\theta cos\theta +a^{2}si{n}^2\theta +b^{2}co{s}^2\theta -2absin\theta cos\theta$
$=a^{2}co{s}^2\theta +b^{2}si{n}^2\theta +a^{2}si{n}^2\theta +b^{2}co{s}^2\theta$
$=a^2(co{s}^2\theta +si{n}^2\theta )+b^2(co{s}^2\theta +si{n}^2\theta )$
$=a^2+b^2$