Consider the equation $\frac{Si{n}^{4}x}{a}+\frac{Co{s}^{4}x}{b}=\frac{1}{a+b},0<x<\frac{\pi }{2}$ then $\frac{Si{n}^{18}x}{a^7}+\frac{Co{s}^{18}x}{b^7}=$

$\left(\frac{a+b}{a}\right)Si{n}^{4}x+\left(\frac{a+b}{b}\right)Co{s}^{4}x=1$

$$\begin{gathered} Si{n}^{4}x+\frac{b}{a}aSi{n}^{4}x+\frac{a}{b}Co{s}^{4}x+Co{s}^{4}x=1 \\ {(Si{n}^{2}x+Co{s}^{2}x)}^2-2Si{n}^{2}x{\cos }^{2}x+\frac{b}{a}Si{n}^{4}x+\frac{a}{b}Co{s}^{4}x=1 \\ \Rightarrow \sqrt{\frac{b}{a}}Si{n}^{2}x-\sqrt{\frac{a}{b}}Co{s}^{2}x=0\Rightarrow {\tan }^{2}x=\frac{a}{b}\Rightarrow Sinx=\frac{\sqrt{a}}{\sqrt{a}+b}Cosx=\frac{\sqrt{b}}{\sqrt{a+b}} \end{gathered}$$

Now,

${\mathrm{Sin}}^{18}x=\frac{a^9}{(a+b{)}^9},{\mathrm{Cos}}^{18}x=\frac{b^9}{(a+b{)}^9}$

$\frac{{\mathrm{Sin}}^{18x}}{a^7}+\frac{{\mathrm{Cos}}^{18}x}{b^7}=\frac{a^2+b^2}{(a+b{)}^9}$