$\sec\left(x\right)=\frac{\cos\left(x\right)}{1-\sin^2\left(x\right)}$
$\tan^2y+6=\sec^2y+5$
$\lim_{x\to\infty}\left(x-\ln\left(e^x+x-1\right)\right)$
$\frac{1}{\sqrt{7+8x^2}}$
$\left(3x-\sqrt{6}\right)^2$
$y^2-9y+8$
$96\left(8\cdot27\right)^2$
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