tanθ=y/x হলে, xcos2θ+ysin2θ এর মান-

Explanation:

Another Explanation (5): bài 📝: \(tan\theta = \frac{y}{x}\) হলে, \(x\cos2\theta + y\sin2\theta\) এর মান নির্ণয় করো। সমাধান: আমরা জানি, \(\cos2\theta = \frac{1 - tan^2\theta}{1 + tan^2\theta}\) এবং \(\sin2\theta = \frac{2tan\theta}{1 + tan^2\theta}\) যেহেতু \(tan\theta = \frac{y}{x}\), সুতরাং \(\cos2\theta = \frac{1 - (\frac{y}{x})^2}{1 + (\frac{y}{x})^2} = \frac{1 - \frac{y^2}{x^2}}{1 + \frac{y^2}{x^2}} = \frac{\frac{x^2 - y^2}{x^2}}{\frac{x^2 + y^2}{x^2}} = \frac{x^2 - y^2}{x^2 + y^2}\) এবং \(\sin2\theta = \frac{2(\frac{y}{x})}{1 + (\frac{y}{x})^2} = \frac{\frac{2y}{x}}{1 + \frac{y^2}{x^2}} = \frac{\frac{2y}{x}}{\frac{x^2 + y^2}{x^2}} = \frac{2xy}{x^2 + y^2}\) এখন, \(x\cos2\theta + y\sin2\theta = x(\frac{x^2 - y^2}{x^2 + y^2}) + y(\frac{2xy}{x^2 + y^2})\) \( = \frac{x(x^2 - y^2) + y(2xy)}{x^2 + y^2} = \frac{x^3 - xy^2 + 2xy^2}{x^2 + y^2} = \frac{x^3 + xy^2}{x^2 + y^2}\) \( = \frac{x(x^2 + y^2)}{x^2 + y^2} = x\) অতএব, \(x\cos2\theta + y\sin2\theta = x\) 🥳 উত্তর: x