$\frac{\cos}{\sec}\cdot\tan^2=\sin^2$
$20x^2+10x$
$\pi-cos^{-1}\left(\frac{10}{26}\right)$
$\left(2x^2-5x+6\right)\left(3x^4-5x^3-6x^2-3\right)$
$3a+2b-c+2a+3b+c$
$\left(2-x\right)\left(2+6x\right)$
$-x^2+2x+10=x+4$
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