cotɑ+cotß=a, tanɑ+tanẞ=b এবং ɑ+ẞ=θ তবে,cot θ এর মান কত?

দেওয়া আছে,

\(\cot \alpha + \cot \beta = a\)

\(\tan \alpha + \tan \beta = b\)

এবং \(\alpha + \beta = \theta\)

আমরা জানি, \(\cot \alpha = \frac{1}{\tan \alpha}\) এবং \(\cot \beta = \frac{1}{\tan \beta}\)

সুতরাং, \(\cot \alpha + \cot \beta = \frac{1}{\tan \alpha} + \frac{1}{\tan \beta} = \frac{\tan \alpha + \tan \beta}{\tan \alpha \tan \beta}\)

অতএব, \(a = \frac{b}{\tan \alpha \tan \beta}\)

সুতরাং, \(\tan \alpha \tan \beta = \frac{b}{a}\)

আবার, \(\cot \theta = \cot (\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}\)

\(\cot \theta = \frac{\frac{1}{\tan \alpha \tan \beta} - 1}{a} = \frac{\frac{1}{\frac{b}{a}} - 1}{a} = \frac{\frac{a}{b} - 1}{a} = \frac{\frac{a-b}{b}}{a} = \frac{a-b}{ab} = \frac{a}{ab} - \frac{b}{ab} = \frac{1}{b} - \frac{1}{a}\)

অতএব, \(\cot \theta = \frac{1}{b} - \frac{1}{a}\) 🥳