Explanation

(a)
Given, $\cot{(\alpha + \beta) = 0\ \Rightarrow \ {cos\ }(\alpha + \beta)} = 0$
$\Rightarrow \alpha + \beta = (2n + 1)\frac{\pi}{2}\ ,\ n \in I$
$\therefore\ \sin{(\alpha + 2\beta) = \sin(2\alpha + 2\beta - \alpha)}$
$= \sin{\lbrack(2n + 1)\pi - \alpha\rbrack}$
$= \sin{(2n\pi + \pi - \alpha)}$
$= \sin{(\pi - \alpha) = \sin\alpha}$