Let x = a sin α θ cos α + 1 θ ,   y = a sin α + 1 θ cos α θ ,…

It is given that $x=a{\sin }^a\theta {\cos }^{a+1}\theta$ and $y=a{\sin }^{a+1}\theta {\cos }^a\theta$

Now, $\frac{{\left(x^2+y^2\right)}^m}{(xy{)}^n}$

Substitute the values in the above expression.

$=\frac{{\left\{{\left(a{\sin }^a\theta {\cos }^{a+1}\theta \right)}^2+{\left(a{\sin }^{a+1}\theta {\cos }^a\theta \right)}^2\right\}}^m}{{\left(a{\sin }^a\theta {\cos }^{\alpha +1}\theta ·a{\sin }^{a+1}\theta {\cos }^a\theta \right)}^n}$

Simplify the above equation.

$x=\frac{a^{2m}{\left({\sin }^{2a}\theta {\cos }^{2a}\theta \right)}^m{\left({\sin }^2\theta +{\cos }^2\theta \right)}^m}{a^{2n}{\left({\sin }^{2a+1}\theta {\cos }^{2a+1}\theta \right)}^n}$

$=\frac{a^{2m-2n}{\left({\sin }^{2a}\theta {\cos }^{2a}\theta \right)}^{m-n}}{(\sin \theta \cos \theta {)}^n}$

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

For independent value of $\theta$

$2\alpha (m-n)-n=0$

$\Rightarrow 2\alpha m-2\alpha n=n$

$\Rightarrow 2\alpha m=n(2\alpha +1)$