Question

true

Answer 1

$$
\cfrac{\csc6x+\cot6x}{\csc6x-\cot6x}=\cot^23x\\
$$
Put everything in terms of sin and cos and eliminate the denominator. Then use the power-reduction formulas to eliminate the square terms:
$$
\csc6x+\cot6x=\cot^23x(\csc6x-\cot6x)\\
\cfrac{1}{\sin6x}+\cfrac{\cos6x}{\sin6x}=\left(\cfrac{\cos3x}{\sin3x}\right)^2\left(\cfrac{1}{\sin6x}-\cfrac{\cos6x}{\sin6x}\right )\\
\cfrac{1+\cos6x}{\sin6x}=\left(\cfrac{\cos3x}{\sin3x}\right)^2\left(\cfrac{1- \cos6x}{\sin6x}\right)\\
1+\cos6x=\left(\cfrac{\cos3x}{\sin3x}\right)^2(1- \cos6x)\\
\sin^23x(1+\cos6x)=\cos^23x(1- \cos6x)\\
\left(\cfrac{1-\cos6x}{\sin6x}\right)(1+\cos6x)=\left(\cfrac{1+\cos6x}{\sin6x}\right)(1- \cos6x)\\
$$
The left-hand side is equal to the right-hand side. Therefore,
$$
\cfrac{\csc6x+\cot6x}{\csc6x-\cot6x}=\cot^23x\\
$$

Any questions?

Answer 2

answer was tan^2 3x :/