Answer
$\frac{tan~x-cot~x}{tan~x+cot~x} = 2sin^2~x-1$
Work Step by Step
$\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{\frac{sin~x}{cos~x}-\frac{cos~x}{sin~x}}{\frac{sin~x}{cos~x}+\frac{cos~x}{sin~x}}$
$\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{\frac{sin^2~x-cos^2~x}{sin~x~cos~x}}{\frac{sin^2~x+cos^2~x}{sin~x~cos~x}}$
$\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{sin^2~x-cos^2~x}{sin^2~x+cos^2~x}$
$\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{sin^2~x-cos^2~x}{1}$
$\frac{tan~x-cot~x}{tan~x+cot~x} = sin^2~x-(1-sin^2~x)$
$\frac{tan~x-cot~x}{tan~x+cot~x} = 2sin^2~x-1$
