$\tan \alpha-\tan \beta=x$ এবং $\cot \beta-\cot \alpha=y$ হলে নিচের কোনটি সঠিক, $\cot (\alpha-\beta)= ?$

Solve: দেওয়া আছে, $\tan \alpha-\tan \beta=x$ এবং $\cot \beta-\cot \alpha=y$এখন; $\frac{1}{x}+\frac{1}{y}=\frac{1}{\tan \alpha-\tan \beta}+\frac{1}{\cot \beta-\cot \alpha}$

$\begin{array}{l} =\frac{1}{\frac{1}{\cot \alpha}-\frac{1}{\cot \beta}}+\frac{1}{\cot \beta-\cot \alpha} \\ =\frac{\cot \alpha \cot \beta}{\cot \beta-\cot \alpha}+\frac{1}{\cot \beta-\cot \alpha} \\ =\frac{\cot \alpha \cot \beta+1}{\cot \beta-\cot \alpha}=\cot (\alpha-\beta) \\ \end{array}$

$\therefore \cot (\alpha-\beta)=\frac{1}{x}+\frac{1}{y}$ (Showed)