দেওয়া আছে, $a = \sin\theta - \sin\phi$ এবং $b = \cos\theta - \cos\phi$। $$ \begin{aligned} \frac{a}{b} &= \frac{\sin\theta - \sin\phi}{\cos\theta - \cos\phi} = \frac{2\cos\frac{\theta+\phi}{2}\sin\frac{\theta-\phi}{2}}{-2\sin\frac{\theta+\phi}{2}\sin\frac{\theta-\phi}{2}} = -\cot\frac{\theta+\phi}{2} \end{aligned} $$ সুতরাং, $\tan\frac{\theta+\phi}{2} = -\frac{b}{a}$। আমরা জানি, $$ \begin{aligned} \cos(\theta+\phi) &= \frac{1 - \tan^2\frac{\theta+\phi}{2}}{1 + \tan^2\frac{\theta+\phi}{2}} \end{aligned} $$ $$ \begin{aligned} &= \frac{1 - (-\frac{b}{a})^2}{1 + (-\frac{b}{a})^2} = \frac{1 - \frac{b^2}{a^2}}{1 + \frac{b^2}{a^2}} = \frac{a^2 - b^2}{a^2 + b^2} \end{aligned} $$