$\left(tan\:x-1\right)^2$

$\left(3x\:-4y\right)\:\left(3x\:+4y\right)$

$tan^2x=\frac{\left(1+cosx\right)\left(secx-1\right)}{cosx}$

$\left(4a^5b^3c\right)\left(8ab^5c^3\right)$

$\int9tan\left(3x\right)ln\left(sec\left(3x\right)\right)dx$

$\lim_{x\to2}\frac{x^3-x^2-3x+2}{x^2-5x+6}$

$6+2^3+1$