How can you change $\tan(x)$ and $\cot(x)$ to $\sin(x)$ and $\cos(x)$?
What ratios using those trig functions do they equal?

Simplify create common denominators in the denominators of the larger fractions.

Simplify each fraction using a Giant One.

Notice you have created a common denominator.
Combine the two fractions.
The numerator is a difference of two squares.
Keep simplifying.

$\frac{\cos(x)}{1-\frac{\text{sin}(x)}{\text{cos}(x)}}-\frac{\sin(x)}{\frac{\text{cos}(x)}{\sin(x)}-1}$

$\frac{\cos \left(x\right)}{\frac{\cos \left(x\right)-\sin \left(x\right)}{\cos \left(x\right)}}-\frac{\sin \left(x\right)}{\frac{\cos \left(x\right)-\sin \left(x\right)}{\sin \left(x\right)}}$

$\frac{\cos \left(x\right)}{\cos \left(x\right)}\cdot \frac{\cos \left(x\right)}{\frac{\cos \left(x\right)-\sin \left(x\right)}{\cos \left(x\right)}}-\frac{\sin \left(x\right)}{\frac{\cos \left(x\right)-\sin \left(x\right)}{\sin \left(x\right)}}\cdot \frac{\sin \left(x\right)}{\sin \left(x\right)}$

$\frac{\cos ^2\left(x\right)}{\cos \left(x\right)-\sin \left(x\right)}-\frac{\sin ^2\left(x\right)}{\cos \left(x\right)-\sin \left(x\right)}$