Tan-ning the Sine

As no specific condition is mentioned, I observed that $\alpha = \beta = \frac {\pi} 4 ; \gamma =0$ satisfy the equation. Substituting it in the sine expression, I got the answer. I used the shortcut, but what Krishna means is:

$\Rightarrow \cot^2 \alpha +\cot^2 \beta+ \cot^2 \gamma+ 2 = \cot^2 \alpha\cdot \cot^2 \beta \cdot \cot^2 \gamma$

$\Rightarrow \csc ^2(\alpha)-1+\csc ^2(\beta)-1\csc ^2(\gamma)-1+2 = \\ \csc^2(\alpha)\csc^2(\beta)\csc^2(\gamma) - \sum\csc^2(\alpha)-\sum \csc^2(\alpha)\csc^2(\beta)-1\\\Rightarrow \sum \csc^2(\alpha)\csc^2(\beta) - \csc^2(\alpha)\cdot \csc^2(\beta)\cdot \csc^2(\gamma)=0$

Multiplying both sides by $\sin^2(\alpha)\cdot\sin^2(\beta)\cdot\sin^2(\gamma)$ we get,

$\\ \Rightarrow \sum \sin ^{ 2 } (\alpha )-1= 0.\\ \Rightarrow \sum \sin^2(\alpha) = 1.$