A geometry problem by Abhyudaya Apoorva

= $\frac{sin(90 - 55)cos55 + cos(90 - 55)sin55}{cosec^2(90-80)- tan^2(80)}$
= $\frac{cos^2(55) + sin^2(55)}{sec^2(80) - tan^2(80)}$
= $\frac{1}{1}$
= 1

$\begin{aligned} X & = \frac {\color{#3D99F6}\sin 35^\circ \cos 55^\circ + \cos 35^\circ \sin 55^\circ}{\csc 10^\circ \csc 10^\circ - \tan 80^\circ \tan 80^\circ} & \small \color{#3D99F6} \text{As }\sin A \cos B + \cos A \sin B = \sin (A+B) \\ & = \frac {\color{#3D99F6}\sin 90^\circ}{\dfrac 1{\color{#D61F06}\sin^2 10^\circ} - \dfrac {\sin^2 80^\circ}{\cos^2 80^\circ}} & \small \color{#D61F06} \text{Note that }\sin (90^\circ - x) =\cos x \\ & = \frac {\color{#3D99F6}1}{\dfrac 1{\color{#D61F06}\cos^2 80^\circ} - \dfrac {\sin^2 80^\circ}{\cos^2 80^\circ}} \\ & = \frac 1{\dfrac {\color{#3D99F6}1-\sin^2 80^\circ}{\cos^2 80^\circ}} & \small \color{#3D99F6} \text{As }1 - \sin^2 x = \cos^2 x \\ & = \frac 1{\dfrac {\color{#3D99F6}\cos^2 80^\circ}{\cos^2 80^\circ}} = \boxed{1} \end{aligned}$