$1-\sec\left(x\right)+\tan\left(x\right)$
$\int7x\left(6x^2+3\right)dx$
$\left(x+50\right)^2$
$\left(4x+2y\right)\left(4x+3y\right)$
$3x^2-39x+108=0$
$\frac{3x^2}{x^2+2}$
$1-sec^2x=-tan^2x$
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