An algebra problem by Sandeep Bhardwaj

Expression can be re written as:- $\dfrac{ \left( \cos (-9x) +i \sin (-9x) \right)^{10} \left( \cos 10x + i \sin 10x \right)^{11} }{\left( \cos 10x +i\sin 10x \right)^{20} \left( \cos (-18x) +i \sin (-18x) \right)^{10}}$ Using Euler's Formula , expression can be simplified as:- $\large \begin{aligned} & =\dfrac{\left(e^{-9xi}\right)^{10}\left(e^{10xi}\right)^{11}}{\left(e^{10xi}\right)^{20}\left(e^{-18xi}\right)^{10}}\\&\\&=\dfrac{\left(e^{-90xi}\right)\left(e^{110xi}\right)}{\left(e^{200xi}\right)\left(e^{-180xi}\right)}\\&\\&=\dfrac{\cancel{e^{20xi}}}{\cancel{e^{20xi}}}=\Huge\color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#0C6AC7}{1}}}}}\end{aligned}$