($ x+\alpha)^n + (x+\beta)^n\;=\; [ (cot\theta -1)+1+i ]^n \;+\; [ (cot\theta -1)+1-i ]^n \\ \{ \because x = cot \theta -1 \} \\ \; \\ \; \\ =[ cot\theta +i ]^n \;+\; [ cot\theta -i ]^n \\ \; \\ \; \\ =[ \dfrac{cos\theta}{sin\theta} +i ]^n \;+\; [ \dfrac{cos\theta}{sin\theta} -i ]^n \\ \; \\ \; \\ =[ \dfrac{cos\theta+isin\theta}{sin\theta} ]^n \;+\; [ \dfrac{cos\theta-sin\theta}{sin\theta} ]^n \\ \; \\ \; \\ = \dfrac{(cos\theta+isin\theta)^n}{sin\theta}^n \;+\; \dfrac{(cos\theta-isin\theta)^n}{sin^n\theta} \\ \; \\ \; \\ = \dfrac{cos \; n\theta+isin\;n\theta+cos\;n\theta-isin\;n\theta}{sin^n\theta} \; \; \; \; \ldots \{ By \ DeMoivre's \ theorem \} \\ \; \\ \; \\ = \dfrac{2cos\;n\theta}{sin^n\theta} \\ \; \\ \; \\ = [(1+i)+(1-i)] cos \;n\theta \cdot cosec^n \theta \; \; \; \{ \because 1+i+1-i=2 \} \\ \; \\ \; \\ \therefore (x+\alpha)^n + (x+\beta)^n\;=\; (\alpha+\beta) cosn\theta \; cosec^n\theta \\ $

Hence Proved.