Step 1:
$\alpha$ and $\beta$ are the roots of the equation $x^2-2px+(p^2+q^2)=0$
Solving $x=\large\frac{2p\pm\sqrt{4p^2-4(p^2+q^2)}}{2}$
$\Rightarrow \large\frac{2p\pm i2q}{2}$
$\Rightarrow p\pm iq$
Step 2:
Let $\alpha=p+iq,$ then $\beta=p-iq\Rightarrow \alpha-\beta=2qi$
We also have $\tan\theta=\large\frac{q}{y+p}$
$\Rightarrow y+p=\large\frac{q}{\tan\theta}=\large\frac{q\cos\theta}{\sin\theta}$
Step 3:
Now $\large\frac{(y+\alpha)^n-(y+\beta)^n}{\alpha-\beta}=\large\frac{(y+p+iq)^n-(y+p-iq)^n}{2qi}$
$\qquad\qquad=\large\frac{(\Large\frac{q\cos\theta}{\sin\theta}\normalsize+iq)^n-(\Large\frac{q\cos\theta}{\sin\theta}\normalsize-iq)^n}{2qi}$
$\qquad\qquad=\large\frac{q^n}{\sin^n\theta}\large\frac{(\cos \theta+i\sin\theta)^n-(\cos\theta-i\sin\theta)^n}{2qi}$
$\qquad\qquad=\large\frac{q^n}{\sin^n\theta}\large\frac{\cos n\theta+i\sin n\theta-\cos n\theta+i\sin n\theta}{2qi}$
$\qquad\qquad=\large\frac{q^n}{2qi\sin^n\theta}$$2i\sin n\theta$
$\qquad\qquad=\large\frac{q^n}{q\sin^n\theta}$$\sin n\theta$
$\qquad\qquad=\large\frac{q^{n-1}}{\sin^n\theta}$$\sin n\theta$
Hence proved.