# How do you verify cot^4 x+ cot^2x = csc^4 x - csc^2 x?

It is

Write ${\csc}^{4} x$ as $\frac{1}{{\sin}^{4} x}$ and ${\cot}^{4} x$ as $\frac{{\cos}^{4} x}{{\sin}^{4} x}$
=> csc^4x - cot^4x = 1/(sin^4 x) - (cos^4 x)/(sin^4 x) = (1 - cos^4 x)/(sin^4 x)

Now recall that ${a}^{2} - {b}^{2} = \left(a - b\right) \cdot \left(a + b\right)$ and use this fact with ${a}^{2}$ being and
${b}^{2}$ being ${\cos}^{4} x$ so that $a$ is 1 and b is ${\cos}^{2} x$

So
${\csc}^{4} x - {\cot}^{4} x = \frac{1 - {\cos}^{4} x}{{\sin}^{4} x} = \left(1 - {\cos}^{2} x\right) \cdot \frac{1 + {\cos}^{2} x}{\sin} ^ 4 x$

But ${\cos}^{2} x + {\sin}^{2} x = 1$ so that $1 - {\cos}^{2} x = {\sin}^{2} x$
so csc^4x - cot^4x = sin^2 x * (1 + cos^2 x) / sin^4 x = (1 + cos^2 x) / sin^2 x = (1/sin^2 x) + (cos^2 x / sin^2 x) = csx^2 x + cos^2 x