Question

If $x\cos \theta -y\sin \theta =a,x\sin \theta +y\cos \theta =b$, prove that $x^2+y^2=a^2+b^2$

Answer 1

$x\cos \theta -y\sin \theta =a$......(1)

$x\sin \theta +y\cos \theta =b$.....(2)

$(1{)}^2+(2{)}^2$

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