$\frac{\left(8sin\left(x\right)+tan\left(x\right)\right)}{tan\left(x\right)}$
$\frac{x^4-5x^3+x^2+5x-2}{x+1}$
$\frac{2x+1}{-4x+8}$
$\frac{6}{x+1}=\frac{7}{x-1}$
$-8+5+3q+6-4q+8q-4q+7$
$\frac{sin4x-sin2x}{2cos3x}=-1$
$\lim_{x\to0}\left(\frac{7\:cos\left(0.01\right)-7\:}{0.01}\right)$
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