Prove the trigonometric identity $\frac{\left(\sec\left(x\right)-\tan\left(x\right)\right)^2+1}{\csc\left(x\right)\left(\sec\left(x\right)-\tan\left(x\right)\right)}=2\tan\left(x\right)$

$\left(\frac{x^{\left(-1\right)\cdot\frac{1}{2}}}{x^{\frac{3}{2}}}\right)^{\frac{2}{5}}$

$\sqrt{\frac{x^3-y^3}{x+y}\cdot\frac{x^2+2x\cdot y+y^2}{x^2+xy+y^2}}\frac{x^2-y^2}{4}$

$\log_{5}\left(5^2\right) x^8$

$5\frac{x^2+3x+5}{2x-1}\cdot\frac{2x^2-2x-13}{4x^2-4x+1}$

$\frac{x^2-6x+9}{4x^2-1}\cdot \frac{8x^3-1}{x^2+5x-24}$

$5\left(\frac{x^2+3x+5}{2x-1}\right)^4\cdot\frac{2x^2-2x-13}{4x^2-4x+1}$

$\log\left(3\right)\cdot x+\log\left(27\right)\cdot x$