How to prove $n$ is even in this trigonometric equation?

Question

I have this trigonometric equation-

$$\bigg(\frac {\cos a+\cos b}{\sin a-\sin b}\bigg)^n+\bigg(\frac {\sin a+\sin b}{\cos a-\cos b}\bigg)^n=2\cot^n \frac {a-b}{2}$$.

How do I prove $n$ is even in this?

Answer 1

We know that $$\cos a+ \cos b = 2\cos \frac {a+b}{2}\cos \frac {a-b}{2} \tag {1}$$ $$\sin a- \sin b = 2\cos \frac {a+b}{2}\sin \frac {a-b}{2} \tag {2} $$ $$\sin a + \sin b = 2\sin \frac {a+b}{2}\cos \frac {a-b}{2} \tag {3}$$ $$\cos a - \cos b = 2\sin \frac {a+b}{2}\sin \frac {a-b}{2}\tag {4} $$

Hope you can take it from here.

Answer 2

\begin{align*} \sin^2 a+\cos^2 a &= \sin^2 b+\cos^2 b \\ \cos^2 a-\cos^2 b &= \sin^2 b-\sin^2 a \\ (\cos a+\cos b)(\cos a-\cos b) &= (\sin a+\sin b)(\sin a-\sin b) \\ \frac{\cos a+\cos b}{\sin a-\sin b} &= \frac{\sin a+\sin b}{\cos b-\cos a} \\ &= \frac{2\sin \frac{a+b}{2} \cos \frac{a-b}{2}} {2\sin \frac{a+b}{2} \sin \frac{a-b}{2}} \\ &= \cot \frac{a-b}{2} \\ \left( \frac{\cos a+\cos b}{\sin a-\sin b} \right)^{n}+ \left( \frac{\sin a+\sin b}{\cos a-\cos b} \right)^{n} &= \left( \cot \frac{a-b}{2} \right)^{n}+ \left( -\cot \frac{a-b}{2} \right)^{n} \end{align*}