Question

If $a\cos \theta +b\sin \theta =m$ and $a\sin \theta -b\cos \theta =n,$ then the value of $a^2+b^2$ is

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

Answer 1

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

Given : $a\cos \theta +b\sin \theta =m$ ---( 1 )

$a\sin \theta -b\cos \theta =n$ ---( 2 )

Squaring (1) and (2), we get

$a^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab\sin \theta \cos \theta =m^2$ ---(3)

and

$a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\sin \theta \cos \theta =n^2$ ---(4)

Adding (3) and (4)

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

or

$a^2+b^2=m^2+n^2$

Hence option ${}^{\prime }{D}^{\prime }$ is the answer.