Question

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

Answer 1

$acos\theta -bsin\theta =x$

Squaring on both sides

$(acos\theta -bsin\theta {)}^2=x^2$

$a^{2}co{s}^2\theta -2abcos\theta sin\theta +b^{2}si{n}^2\theta =x^2......................\left(1\right)$

$asin\theta +bcos\theta =y$

$(asin\theta +bcos\theta {)}^2=y^2$

$a^{2}si{n}^2\theta +2absin\theta cos\theta +b^{2}co{s}^2\theta =y^2.....................\left(2\right)$

Adding (1) and(2)

$a^{2}co{s}^2\theta -2absin\theta cos\theta +b^{2}si{n}^2\theta$+$a^{2}si{n}^2\theta +2absin\theta cos\theta +b^{2}co{s}^2\theta =x^2+y^2$

$a^2(si{n}^2\theta +co{s}^2\theta )+b^2(si{n}^2\theta +co{s}^2\theta )=x^2+y^2$

$si{n}^2\theta +co{s}^2\theta =1$

$a^2+b^2=x^2+y^2$