Question

If $a\cos \theta -b\sin \theta =x$ and $a\sin \theta +b\cos \theta =y$ then find $\frac{a^2+b^2}{x^2+y^2}$.

Answer 1

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

Now, $x^2+y^2=a^{2}co{s}^2\theta +b^{2}si{n}^2\theta -2abcos\theta +a^{2}si{n}^2\theta +b^{2}co{s}^2\theta +2abcos\theta sin\theta =a^2(co{s}^2\theta +si{n}^2\theta )+b^2(co{s}^2\theta +si{n}^2\theta )=a^2+b^2$

Hence, $\frac{a^2+b^2}{x^2+y^2}=\frac{a^2+b^2}{a^2+b^2}=1$