$\cos\left(-\frac{5}{3}\right)\cdot\cos\left(\frac{4}{5}\right)-\sin\left(-\frac{5}{3}\right)\cdot\sin\left(\frac{4}{5}\right)$

$y'\:-\frac{2y}{x+1}=\left(x+1\right)^3$

$y=y+4$

$\frac{cos\left(x\right)}{1+sin\left(x\right)}-\frac{1}{cos\left(x\right)}=-\tan\left(x\right)$

$\lim_{x\to3}\left(\frac{2x^2+\:12x-\:18}{x^2+\:x-6}\right)$

$\cos^2x+4\cos x=0$

$\left(s+4\right)^2$