$\frac{sec\:x\:-1}{tan\:x}=csc\:x\:-\:cot\:x$
$\left(2a^3b^4\right)\cdot b^3$
$\frac{6x^3+4x^2+6x+7}{x-5}$
$1\sin x=\frac{\cos^2\left(x\right)}{1+\sin\left(x\right)}$
$9^2\:x\:\:9^9$
$e^{9z}=xyz$
$\left(8-y\right)^3$
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