tanθ+sinθ=m, tanθ-sinθ=n হলে m²-n²=?

দেওয়া আছে, \(tan\theta + sin\theta = m\) এবং \(tan\theta - sin\theta = n\)

তাহলে, \(m^2 - n^2 = (tan\theta + sin\theta)^2 - (tan\theta - sin\theta)^2\)

আমরা জানি, \((a+b)^2 - (a-b)^2 = 4ab\)

সুতরাং, \(m^2 - n^2 = 4 \cdot tan\theta \cdot sin\theta\)

এখন, \(mn = (tan\theta + sin\theta)(tan\theta - sin\theta) = tan^2\theta - sin^2\theta\)

\(= \frac{sin^2\theta}{cos^2\theta} - sin^2\theta = sin^2\theta (\frac{1}{cos^2\theta} - 1) = sin^2\theta (\frac{1-cos^2\theta}{cos^2\theta})\)

\(= sin^2\theta \cdot \frac{sin^2\theta}{cos^2\theta} = sin^2\theta \cdot tan^2\theta\)

সুতরাং, \(mn = sin^2\theta \cdot tan^2\theta\)

অতএব, \(\sqrt{mn} = \sqrt{sin^2\theta \cdot tan^2\theta} = sin\theta \cdot tan\theta\)

সুতরাং, \(4\sqrt{mn} = 4 \cdot sin\theta \cdot tan\theta\)

সুতরাং, \(m^2 - n^2 = 4tan\theta sin\theta = 4\sqrt{mn}\). 🎉