যদি  tantheta=y/x  হয়,  তবে  x cos 2theta + ysin2theta  এর মান হবে -- 

Explanation:

Another Explanation (5): দেওয়া আছে, \( \tan\theta = \frac{y}{x} \) আমাদের \( x \cos 2\theta + y \sin 2\theta \) এর মান নির্ণয় করতে হবে। আমরা জানি, \( \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \) এবং \( \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \) এখন, \( x \cos 2\theta + y \sin 2\theta = x \cdot \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} + y \cdot \frac{2 \tan \theta}{1 + \tan^2 \theta} \) \( \tan \theta = \frac{y}{x} \) বসিয়ে পাই, \( = x \cdot \frac{1 - \frac{y^2}{x^2}}{1 + \frac{y^2}{x^2}} + y \cdot \frac{2 \cdot \frac{y}{x}}{1 + \frac{y^2}{x^2}} \) \( = x \cdot \frac{\frac{x^2 - y^2}{x^2}}{\frac{x^2 + y^2}{x^2}} + y \cdot \frac{\frac{2y}{x}}{\frac{x^2 + y^2}{x^2}} \) \( = x \cdot \frac{x^2 - y^2}{x^2 + y^2} + y \cdot \frac{2yx}{x^2 + y^2} \) \( = \frac{x(x^2 - y^2) + 2y^2x}{x^2 + y^2} \) \( = \frac{x^3 - xy^2 + 2xy^2}{x^2 + y^2} \) \( = \frac{x^3 + xy^2}{x^2 + y^2} \) \( = \frac{x(x^2 + y^2)}{x^2 + y^2} \) \( = x \) সুতরাং, \( x \cos 2\theta + y \sin 2\theta = x \) 🥳🎉