Question

Sec^6x (secxtanx) - sec^4x (secxtanx) = sec^5x tan^3x

Answer 1

Let E the expression of the left side of the equation:

$E=\sec^6(x) (\sec (x)\tan (x)) - \sec^4(x) (\sec (x)\tan (x))\\\\ E=\sec^7(x)\tan(x)-\sec^5(x)\tan(x)\\\\ E=\sec^5(x)\tan(x)(\sec^2(x)-1)$

We must know that $\tan^2(x)+1=\sec^2(x)\iff \sec^2(x)-1=\tan^2(x)$. So, using in the equation above:

$E=\sec^5(x)\tan(x)(\underbrace{\sec^2(x)-1}_{\tan^2(x)})\\\\ E=\sec^5(x)\tan(x)(\tan^2(x))\\\\ \boxed{E=\sec^5(x)\tan^3(x)}$

Then, the equation is true.