Explanation:

Another Explanation (5): ```html

সমাধান:

দেওয়া আছে, \( tan\theta = \frac{y}{x} \)। আমাদের \( x\cos2\theta + y\sin2\theta \) এর মান নির্ণয় করতে হবে। আমরা জানি, \[ \cos2\theta = \frac{1 - tan^2\theta}{1 + tan^2\theta} \] এবং \[ \sin2\theta = \frac{2tan\theta}{1 + tan^2\theta} \] এখন, \( tan\theta = \frac{y}{x} \) এই মানগুলো বসালে পাই, \[ \cos2\theta = \frac{1 - (\frac{y}{x})^2}{1 + (\frac{y}{x})^2} = \frac{1 - \frac{y^2}{x^2}}{1 + \frac{y^2}{x^2}} = \frac{\frac{x^2 - y^2}{x^2}}{\frac{x^2 + y^2}{x^2}} = \frac{x^2 - y^2}{x^2 + y^2} \] \[ \sin2\theta = \frac{2(\frac{y}{x})}{1 + (\frac{y}{x})^2} = \frac{\frac{2y}{x}}{1 + \frac{y^2}{x^2}} = \frac{\frac{2y}{x}}{\frac{x^2 + y^2}{x^2}} = \frac{2xy}{x^2 + y^2} \] সুতরাং, \( x\cos2\theta + y\sin2\theta = x(\frac{x^2 - y^2}{x^2 + y^2}) + y(\frac{2xy}{x^2 + y^2}) \) \( = \frac{x(x^2 - y^2) + y(2xy)}{x^2 + y^2} \) \( = \frac{x^3 - xy^2 + 2xy^2}{x^2 + y^2} \) \( = \frac{x^3 + xy^2}{x^2 + y^2} \) \( = \frac{x(x^2 + y^2)}{x^2 + y^2} \) \( = x \) অতএব, \( x\cos2\theta + y\sin2\theta = x \) 🥳 ```