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**Soroban** Hello, vulcangon!

Welcome abaord!

$\displaystyle \frac{\sin^2\!x - \tan x}{\cos^2\!x - \cot x} \;=\;\frac{\sin^2\!x - \dfrac{\sin x}{\cos x}}{\cos^2\!x - \dfrac{\cos x}{\sin x}} $

. . . . . . . . $\displaystyle = \;\frac{\sin x\left(\sin x - \dfrac{1}{\cos x}\right)} {\cos x\left(\cos x - \dfrac{1}{\sin x}\right)} $

. . . . . . . . $\displaystyle =\; \frac{\sin x\left(\dfrac{\sin x\cos x - 1}{\cos x}\right)} {\cos x\left(\dfrac{\sin x\cos x - 1}{\sin x}\right)} $

. . . . . . . . $\displaystyle =\; \frac{\dfrac{\sin x}{\cos x}(\sin x\cos x - 1)}{\dfrac{\cos x}{\sin x}(\sin x\cos x - 1)} $

. . . . . . . . $\displaystyle = \;\frac{\tan x}{\cot x}$

. . . . . . . . $\displaystyle =\; \frac{\tan x}{\frac{1}{\tan x }}$

. . . . . . . . $\displaystyle =\;\tan^2\!x$