$\theta$ কোণকে $\alpha$ ও $\beta$ অংশে এমনভাবে বিভক্ত করা হলো যেন $\tan\ \alpha\ =\ \text{k}\tan\ \beta$ হয়, প্রমাণ কর যে, $\sin\ \left(\alpha\ -\ \beta\right)=\frac{k-1}{k+1}\sin\theta$

$\begin{array}{l} \tan \alpha=\mathrm{k} \tan \beta \Rightarrow \frac{\sin \alpha}{\cos \alpha}=\mathrm{k} \frac{\sin \beta}{\cos \beta} \Rightarrow \frac{\sin \alpha \cos \beta}{\cos \alpha \sin \beta}=\frac{\mathrm{k}}{1} \\ \Rightarrow \frac{\sin \alpha \cos \beta-\cos \alpha \sin \beta}{\sin \alpha \cos \beta+\cos \alpha \sin \beta}=\frac{\mathrm{k}-1}{\mathrm{k}+1} \quad \Rightarrow \frac{\sin (\alpha-\beta)}{\sin (\alpha+\beta)}=\frac{\mathrm{k}-1}{\mathrm{k}+1} \\ \Rightarrow \sin (\alpha-\beta)=\frac{\mathrm{k}-1}{\mathrm{k}+1} \sin \theta \quad[\alpha+\beta=\theta] \quad \text { (Proved) } \end{array}$