tanθ = y/x হলে x cos2θ + y sin2θ =? 

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 \( \sin2\theta = \frac{2\tan\theta}{1 + \tan^2\theta} \).

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,

\(\sin2\theta = \frac{2(\frac{y}{x})}{1 + (\frac{y}{x})^2} = \frac{\frac{2y}{x}}{1 + \frac{y^2}{x^2}} = \frac{\frac{2y}{x}}{\frac{x^2 + y^2}{x^2}} = \frac{2xy}{x^2 + y^2}\).

অতএব,

\( x \cos2\theta + y \sin2\theta = x \cdot \frac{x^2 - y^2}{x^2 + y^2} + y \cdot \frac{2xy}{x^2 + y^2} \)

\(= \frac{x(x^2 - y^2) + y(2xy)}{x^2 + y^2} \)

\(= \frac{x^3 - xy^2 + 2xy^2}{x^2 + y^2} \)

\(= \frac{x^3 + xy^2}{x^2 + y^2} \)

\(= \frac{x(x^2 + y^2)}{x^2 + y^2} \)

\(= x \). 🎉

সুতরাং, \( x \cos2\theta + y \sin2\theta = x \). 🥳