$\sec\left(x\right)+\cos\left(x\right)=\frac{1}{\cos\left(x\right)}+\frac{1}{\sec^2\left(x\right)}$
$\sqrt{16}^2$
$2\sqrt{40}-\sqrt{60}$
$y^1+4x=5$
$\frac{-12a^4b^2}{-3a^2b}$
$f ( x ) = 2 x - 2$
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