If α + β − γ = π

Given:

$\gamma=-[\pi-(\alpha+\beta)]$

Also,

$\lambda=\frac{\sin ^{2} \alpha+\sin ^{2} \beta-\sin ^{2}[-(\pi-(\alpha+\beta)]}{\sin \alpha \sin \beta \cos (-(\pi-(\alpha+\beta))}$

$=\frac{\sin ^{2} \alpha+\sin ^{2} \beta-(\sin (\alpha+\beta))^{2}}{-(\sin \alpha \sin \beta \cos (\alpha+\beta))} \quad[\sin (\pi-\theta)=\sin \theta$ and $\cos (\pi-\theta)=-\cos \theta]$

$=\frac{\sin ^{2} \alpha+\sin ^{2} \beta-\sin ^{2} \alpha \cos ^{2} \beta-\cos ^{2} \alpha \sin ^{2} \beta-2 \sin \alpha \sin \beta \cos \alpha \cos \beta}{-\left(\sin \alpha \sin \beta \cos \alpha \cos \beta-\sin ^{2} \alpha \sin ^{2} \beta\right)}$

$=\frac{\sin ^{2} \alpha\left(1-\cos ^{2} \beta\right)+\sin ^{2} \beta\left(1-\cos ^{2} \alpha\right)-2 \sin \alpha \sin \beta \cos \alpha \cos \beta}{\sin ^{2} \alpha \sin ^{2} \beta-\sin \alpha \sin \beta \cos \alpha \cos \beta}$

$=\frac{2 \sin ^{2} \alpha \sin ^{2} \beta-2 \sin \alpha \sin \beta \cos \alpha \cos \beta}{\sin ^{2} \alpha \sin ^{2} \beta-\sin \alpha \sin \beta \cos \alpha \cos \beta}$

$=2$