# Find $\lambda$ if $\alpha + \beta - \gamma = \pi$, and $\sin ^ 2\alpha + \sin^2\beta - \sin^2\gamma = \lambda\sin\alpha\sin\beta\cos\gamma$

If $\alpha + \beta - \gamma = \pi$, and $\sin ^ 2\alpha + \sin^2\beta - \sin^2\gamma = \lambda\sin\alpha\sin\beta\cos\gamma$, then find $\lambda$.

From the first equation we get that $\sin^2\gamma = sin^2(\alpha+\beta) \text{ and } \cos\gamma = -(\alpha+\beta)$.
Putting this in the second equation, we get

$$(2-\lambda)(\tan\alpha\tan\beta - 1) = 0$$
So $\lambda =\boxed{2}$
However, the procedure that I did is very long. Is there a simpler way to solve it?

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$$\sin^2\beta-\sin^2\gamma=\sin(\beta-\gamma)\sin(\beta+\gamma)=\sin(\pi-\alpha)\sin(\beta+\gamma)$$

$$\text{As }\sin(\pi-x)=\sin x, \sin^2\beta-\sin^2\gamma=\sin(\alpha)\sin(\beta+\gamma)$$

$$\implies\sin^2\alpha+\sin^2\beta-\sin^2\gamma=\sin^2\alpha+\sin(\alpha)\sin(\beta+\gamma) =\sin\alpha[\sin\alpha+\sin(\beta+\gamma)]$$

$$\text{Again,}\sin\alpha=\sin[\pi-(\beta-\gamma)]=\sin(\beta-\gamma)$$

$$\text{Finally, }\sin(\beta+\gamma)+\sin(\beta-\gamma)=?$$

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