$\left(r\:+\:t\right)\left(s\:+\:m\right)$

$\left(1+tan^2\left(x\right)\right)^2=sec^4\left(x\right)-4sec^2\left(x\right)$

$32x^2+24xy-18y^2$

$tan\:\left(x-5\right)=0.28$

$\frac{3x^4+4x^3+6x-2}{x-5}$

$-\left(-19+3-7\right)-\left\{+5-\left(-7-4\right)\right\}$

$\left(4a^3+5\right)\left(4a^3-5\right)$