$\frac{\tan a}{\sec a}=\sin a$
$\sqrt{15y^3}$
$\left(-\frac{x}{\sqrt{16-x^2}}\right)^2$
$-x^2+6x-5\ge0$
$5\cdot10^3$
$x^2-6x=1$
$\frac{2x+1}{x^2+14x+49}$
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