$\frac{tan^2x+1}{tanx+1}=sec^2x-tanx$
$-\:24\:x\:-5\:x\:-6\:x\:+3$
$\frac{dy}{dx}+\cos\left(x\right)y=3\cos\left(x\right)$
$\frac{sin\:y}{cos\:y}+\frac{cos\:y}{sin\:y}$
$\int\left(senx+cosx-x^2\right)dx$
$\:\frac{2x+6}{4x^2-12x+9}$
$4\sec^2x-8=0$
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