Explanation:

Another Explanation (5): ```html

সমাধান

দেওয়া আছে, \( \tan\theta = \frac{y}{x} \) আমাদের বের করতে হবে: \( x \cos2\theta + y \sin2\theta \) = ? আমরা জানি, \( \cos2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} \) এবং \( \sin2\theta = \frac{2\tan\theta}{1 + \tan^2\theta} \) এখন, \( \tan\theta \) এর মান বসিয়ে পাই, \( \cos2\theta = \frac{1 - (\frac{y}{x})^2}{1 + (\frac{y}{x})^2} = \frac{1 - \frac{y^2}{x^2}}{1 + \frac{y^2}{x^2}} = \frac{\frac{x^2 - y^2}{x^2}}{\frac{x^2 + y^2}{x^2}} = \frac{x^2 - y^2}{x^2 + y^2} \) এবং, \( \sin2\theta = \frac{2(\frac{y}{x})}{1 + (\frac{y}{x})^2} = \frac{\frac{2y}{x}}{1 + \frac{y^2}{x^2}} = \frac{\frac{2y}{x}}{\frac{x^2 + y^2}{x^2}} = \frac{2xy}{x^2 + y^2} \) সুতরাং, \( x \cos2\theta + y \sin2\theta = x \cdot \frac{x^2 - y^2}{x^2 + y^2} + y \cdot \frac{2xy}{x^2 + y^2} \) \( = \frac{x(x^2 - y^2) + y(2xy)}{x^2 + y^2} \) \( = \frac{x^3 - xy^2 + 2xy^2}{x^2 + y^2} \) \( = \frac{x^3 + xy^2}{x^2 + y^2} \) \( = \frac{x(x^2 + y^2)}{x^2 + y^2} \) \( = x \) অতএব, \( x \cos2\theta + y \sin2\theta = x \) 🎉 ```