ধরি, $\lambda = \frac{2 \sin \alpha}{1+\sin \alpha+\cos \alpha}$ এবং $x = \frac{1+\sin \alpha-\cos \alpha}{1+\sin \alpha}$। $$ \begin{aligned} \frac{\lambda}{x} &= \frac{2 \sin \alpha}{1+\sin \alpha+\cos \alpha} \cdot \frac{1+\sin \alpha}{1+\sin \alpha-\cos \alpha} \\ &= \frac{2 \sin \alpha(1+\sin \alpha)}{(1+\sin \alpha)^2 - \cos^2 \alpha} \\ &= \frac{2 \sin \alpha(1+\sin \alpha)}{1+2\sin \alpha+\sin^2 \alpha - \cos^2 \alpha} \\ &= \frac{2 \sin \alpha(1+\sin \alpha)}{1+2\sin \alpha+\sin^2 \alpha - (1-\sin^2 \alpha)} \\ &= \frac{2 \sin \alpha(1+\sin \alpha)}{2\sin \alpha + 2\sin^2 \alpha} \\ &= \frac{2 \sin \alpha(1+\sin \alpha)}{2\sin \alpha(1+\sin \alpha)} = 1 \end{aligned} $$ সুতরাং, $x = \lambda$।