$\frac{{\cot }^2{15}^{\circ }-1}{{\cot }^2{15}^{\circ }+1}=?$

$\frac{1-{\tan }^{2}15}{1+{\tan }^{2}15}=\frac{1-{\left[\tan \left({45}^{\circ }-{30}^{\circ }\right)\right]}^2}{1+{\left[\tan \left({45}^{\circ }-{30}^{\circ }\right)\right]}^2}=\frac{1-\left[\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}\right]}{1+{\left[\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}\right]}^2}$

$\begin{array}{l}=\frac{1-\left[{\left.\frac{\sqrt{3}-1}{\sqrt{3}+1}\right|}^2\right.}{1+{\left[\frac{\sqrt{3}-1}{\sqrt{3}+1}\right]}^2}=\frac{[\sqrt{3}+1{]}^2-[\sqrt{3}-1{]}^2}{[\sqrt{3}+1{]}^2+[\sqrt{3}-1{]}^2}=\frac{4\sqrt{3}}{8}=\frac{\sqrt{3}}{2}\\ \cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\Rightarrow \frac{1-{\tan }^2{15}^{\circ }}{1+{\tan }^2{15}^{\circ }}=\cos {30}^{\circ }=\frac{\sqrt{3}}{2}\end{array}$