$\frac{\left(\sin\left(x\right)\right)^2}{\left(\cos\left(x\right)\right)^3}$
$-7sin^2+4sin=-3$
$9x^2-12+16$
$x\:\cdot\:6x$
$\left(-56\right)+\left(-78\right)-\left(65\right)-\left(1\right)$
$+4\cdot3+4\cdot5$
$2x^2-7x+6$
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