$\alpha$ and $n \alpha$ are two angles and they formed compound angles $(n+1)\alpha$ and $(n-1)\alpha$ by sum and difference. Sine and cosine functions formed a fractional function to represent a quantity in general trigonometric form.
$\dfrac{\sin(n+1)\alpha -\sin(n-1)\alpha}{\cos(n+1)\alpha +2\cos n\alpha +\cos(n-1)\alpha}$
The two sine functions with compound angles are in subtraction form. The subtraction of them can be simplified by the sum to product transformation trigonometric identity.
$= \dfrac{2\cos \Bigg[\dfrac{(n+1)\alpha + (n-1)\alpha}{2}\Bigg] \sin \Bigg[\dfrac{(n+1)\alpha -(n-1)\alpha}{2}\Bigg]}{\cos(n+1)\alpha +\cos(n-1)\alpha +2\cos n\alpha}$
Similarly, express the sum of cosine functions in terms of product form by the sum to product transformation trigonometric identity.
$= \dfrac{2\cos \Bigg[\dfrac{(n+1)\alpha + (n-1)\alpha}{2}\Bigg] \sin \Bigg[\dfrac{(n+1)\alpha -(n-1)\alpha}{2}\Bigg]}{2\cos \Bigg[\dfrac{(n+1)\alpha + (n-1)\alpha}{2}\Bigg] \cos \Bigg[\dfrac{(n+1)\alpha -(n-1)\alpha}{2}\Bigg] +2\cos n\alpha}$
$= \dfrac{2\cos \Bigg[\dfrac{n\alpha +\alpha + n\alpha -\alpha}{2}\Bigg] \sin \Bigg[\dfrac{n\alpha +\alpha -n\alpha +\alpha}{2}\Bigg]}{2\cos \Bigg[\dfrac{n\alpha +\alpha + n\alpha -\alpha}{2}\Bigg] \cos \Bigg[\dfrac{n\alpha +\alpha -n\alpha +\alpha}{2}\Bigg] +2\cos n\alpha}$
$\require{cancel} = \dfrac{2\cos \Bigg[\dfrac{n\alpha +\cancel{\alpha} + n\alpha -\cancel{\alpha}}{2}\Bigg] \sin \Bigg[\dfrac{\cancel{n\alpha} +\alpha -\cancel{n\alpha} +\alpha}{2}\Bigg]}{2\cos \Bigg[\dfrac{n\alpha +\cancel{\alpha} + n\alpha -\cancel{\alpha}}{2}\Bigg] \cos \Bigg[\dfrac{\cancel{n\alpha }+\alpha -\cancel{n\alpha} +\alpha}{2}\Bigg] +2\cos n\alpha}$
$= \dfrac{2\cos \Bigg[\dfrac{n\alpha +n\alpha}{2}\Bigg] \sin \Bigg[\dfrac{\alpha +\alpha}{2}\Bigg]}{2\cos \Bigg[\dfrac{n\alpha + n\alpha}{2}\Bigg] \cos \Bigg[\dfrac{\alpha +\alpha}{2}\Bigg] +2\cos n\alpha}$
$= \dfrac{2\cos \Bigg[\dfrac{2n\alpha}{2}\Bigg] \sin \Bigg[\dfrac{2\alpha}{2}\Bigg]}{2\cos \Bigg[\dfrac{2n\alpha}{2}\Bigg] \cos \Bigg[\dfrac{2\alpha}{2}\Bigg] +2\cos n\alpha}$
$\require{cancel} = \dfrac{2\cos \Bigg[\dfrac{\cancel{2}n\alpha}{\cancel{2}}\Bigg] \sin \Bigg[\dfrac{\cancel{2}\alpha}{\cancel{2}}\Bigg]}{2\cos \Bigg[\dfrac{\cancel{2}n\alpha}{\cancel{2}}\Bigg] \cos \Bigg[\dfrac{\cancel{2}\alpha}{\cancel{2}}\Bigg] +2\cos n\alpha}$
$= \dfrac{2\cos n\alpha \sin \alpha}{2\cos n\alpha \cos \alpha +2\cos n\alpha}$
$\cos n \alpha$ is a common term in expression of the denominator. Take it common from them to simplify it further.
$= \dfrac{2\cos n\alpha \sin \alpha}{2\cos n\alpha (\cos \alpha +1)}$
$\require{cancel} = \dfrac{\cancel{2\cos n\alpha} \sin \alpha}{\cancel{2\cos n\alpha} (\cos \alpha +1)}$
$= \dfrac{\sin \alpha}{\cos \alpha +1}$
$= \dfrac{\sin \alpha}{1+\cos \alpha}$
Expand the sine of angle alpha and also express $1+cos \alpha$.
$= \dfrac{2\sin \Big(\dfrac{\alpha}{2}\Big)\cos \Big(\dfrac{\alpha}{2}\Big) }{2cos^2 \Big(\dfrac{\alpha}{2}\Big)}$
$\require{cancel} = \dfrac{\cancel{2}\sin \Big(\dfrac{\alpha}{2}\Big) \cancel{\cos \Big(\dfrac{\alpha}{2}\Big)} }{\cancel{2cos^2 \Big(\dfrac{\alpha}{2}\Big)}}$
$= \dfrac{\sin \Big(\dfrac{\alpha}{2}\Big)}{cos \Big(\dfrac{\alpha}{2}\Big)}$
$= \tan \Big(\dfrac{\alpha}{2}\Big)$
Therefore, it is the required solution for this trigonometric problem.
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