$secx=-4$
$y\:=\:sin^4x\cdot\frac{tan^6x}{\left(x^2+3\right)^2}$
$\left(8y^2\right)\left(\frac{1}{6}y^{12}\right)^{-1}$
$\sin^2\left(\frac{1}{4}\right)+\cos^2\left(a\right)=1$
$rs+4st$
$x^2+2xy=2y+1$
$\frac{10x^2-17x+3}{2x-3}$
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