दिया गया है, $\cot (\alpha  + \beta ) = 0 \Rightarrow \cos (\alpha  + \beta ) = 0$

$\Rightarrow$ $\alpha  + \beta  = (2n + 1)\frac{\pi }{2},n \in I$

$\therefore$ $\sin (\alpha  + 2\beta ) = \sin (2\alpha  + 2\beta  – \alpha )$

$=\sin {\rm{ }}[(2n + 1){\rm{ }}\pi  – \alpha ]$

$ = \sin (\,2n\pi  + \pi  – \alpha \,)$ = $\sin (\,\pi  – \alpha \,)\, = \sin \alpha $.