$\frac{1}{1+\cos^2}+\frac{1}{1+\sec^2a}$
$\left(\frac{2}{5}x+4\right)^2$
$m^4+m^3+m^2=0$
$17g^2-2g^2$
$5^0+4^2-2^3$
$y'-\left(tan\left(x\right)\right)y=x\left(sec\left(x\right)\right)$
$\:8\left(4z+1\right)+4z$
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