$\frac{12a^5b^2}{6\:ab^5}$
$\frac{\sin x-\sin^3x}{\cos^3x}=\tan x$
$\left(11x+9\right)\left(11x-12\right)$
$2\cdot0.75$
$3tan3x=3$
$\left(\sqrt{2}x+\sqrt{8}\right)^2$
$\frac{sen\:a-sen\:b}{\:cos\:a-cos\:b}+\frac{cos\:a\:+cos\:b}{sen\:a+sen\:b}$
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