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Prove that $n$ is even

$$\left(\dfrac {\cos A + \cos B}{\sin A-\sin B}\right)^n + \left(\dfrac {\sin A+\sin B}{\cos A- \cos B}\right)^n = 2\cot^n \left(\dfrac {A-B}{2}\right)$$

Please help. I didn't get any idea regarding this.

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    Do you mean that if $n$ is even, you want to prove the identity? – 2017-02-17
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    @ m-agag 2016, It is required to prove that $n$ is even.. – 2017-02-17
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    I think he means to use the given equality to show that $n$ is even. – 2017-02-17
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    $A$ and $B$ are any numbers which make the identity sense? – 2017-02-17
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    @ m-agag 2016, $\A $ and $\B $ are angles in degrees – 2017-02-17
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    degree can be changed into radians. – 2017-02-17
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    @ m-agag 2016, I couldn't understand what you mean to say?? – 2017-02-17
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    Use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html – 2017-02-17

2 Answers 2

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Hint:

$$\dfrac {\cos A + \cos B}{\sin A-\sin B} = - \,\dfrac {\sin A+\sin B}{\cos A- \cos B} = \cot \left(\dfrac {A-B}{2}\right)$$

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    @ dxiv, what is this? How does it complete the proof? Could you please elaborate? – 2017-02-17
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    @ÉvaristeGalois First, prove the above using the usual trigonometric identities. Then, $a=-b=c$ and you know that $a^n+b^n=2c^n$ $\iff$ $a^n+(-a)^{n}=2a^n\,$, which should tell you something about $n$. – 2017-02-17
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    Wish the downvoter had left a comment why. – 2017-02-17
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Or, even simplier, one could just choose any value for $A$, and $B$; then use the equality given to deduce that $n$ must be even, like this:

By substituting $A = 0$; and $B = \frac{\pi}{2}$; we'll arrive at: $$\underbrace{(-1)^n}_{\text{swap this one to the other side}} + \underbrace{1^n}_{\text{this is }1} = 2\times(-1)^n$$ Changing sides will yield: $$1 = (-1)^n$$

Which means that if the equality holds for all $A$, and $B$, then we must have $n$ is even.