Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

$\left(\frac{a+b}{a}\right){\sin }^{4}x+\left(\frac{a+b}{b}\right){\cos }^{4}x=1\Rightarrow {\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x+\frac{b}{a}{\sin }^{4}x+\frac{a}{b}{\cos }^{4}x=1\Rightarrow {\left(\sqrt{\frac{b}{a}}{\sin }^{2}x-\sqrt{\frac{a}{b}}{\cos }^{2}x\right)}^2=0\Rightarrow {\tan }^{2}x=\frac{a}{b}.$

So ${\sin }^{2}x=\frac{a}{a+b}$ and ${\cos }^{2}x=\frac{b}{a+b}$. ${\sin }^{18}x=\frac{a^9}{(a+b{)}^9},{\cos }^{18}x=\frac{b^9}{(a+b{)}^9}$, so the sum is $\frac{a^2+b^2}{(a+b{)}^9}$.