$\frac{\sqrt{1+\tan^2\left(x\right)}}{\tan\left(x\right)}$

$25c^2+40cn+16n^2$

$4x-3=17$

$\left(\frac{\left(a^{-2}b^{-3}\right)}{\left(x^{-1}-y^2\right)}\right)^{-3}\left(\frac{\left(x^2b\right)}{\left(a^{\left(-\frac{2}{3}\right)}y^{-3}\right)}\right)$

$\left(2x+3\right)\left(2x-4\right)$

$\frac{20}{9}+\frac{2}{-9}$

$\tan^2\left(1+\cot^2\right)=\frac{1}{1-\sin^2}$