$3\sin\left(x\right)=1$
$6\cdot\left(-7\right)+-9$
$1\:-3\:1\:3$
$\frac{1-\tan^2\left(b\right)}{1+\tan^2\left(b\right)}=1-2\sin^2\left(b\right)$
$\int3z^2dz$
$\left(-2x^2y^3\right)^3$
$\left(\tan a+1\right)\cos^2\:a=1$
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