$\frac{1-\sin x}{1+\sin x}$
$x^2-4x+1^2$
$y'=\frac{4y}{x\left(y-3\right)}$
$\left(2x+4\right)x\left(x-7\right)$
$\left(8xy\:-\:4z\right)\left(8xy\:+\:4z\right)\:$
$\lim_{x\to\infty}\left(\frac{x^2}{1-\cos\left(x\right)}\right)$
$cos\left(2x+50\right)=0.41$
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