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Prove, for angles $\alpha$, $\beta$ and $\gamma$ of a right triangle, the follow

Source: Eotvos 1897 p1

March 14, 2020
geometry

Problem Statement

Prove, for angles α\alpha, β\beta and γ\gamma of a right triangle, the following relation: sin α sin β sin (α−β) + sin β sin γ sin (β−γ) + sin γ sin α sin (γ−α) + sin (α−β) sin (β−γ) sin (γ−α)=0.\text{sin } \alpha \text{ sin } \beta \text{ sin } (\alpha-\beta) \text{ } + \text{ sin } \beta \text{ sin } \gamma \text{ sin } (\beta-\gamma) \text{ }+ \text{ sin } \gamma \text{ sin } \alpha \text{ sin } (\gamma-\alpha) \text{ }+ \text{ sin } (\alpha-\beta) \text{ sin } (\beta-\gamma) \text{ sin } (\gamma-\alpha) = 0.
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