Question

If $x=a\cos \theta -b\sin \theta andy=a\sin \theta +bcos\theta$, then $a^2+b^2=x^2+y^2$.

• A. True
• B. False

Answer 1

The correct option is A True

$x=a\cos \theta -b\sin \theta ,y=a\sin \theta +b\cos \theta$

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

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

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

$=a^2+b^2$ since ${\sin }^2\theta {\cos }^2\theta =1$

Hence $a^2+b^2=x^2+y^2$ is true.