দেওয়া আছে, $\tan \theta + \sec \theta = x$। আমরা জানি, $\sec^2 \theta - \tan^2 \theta = 1$ বা, $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$। সুতরাং, $\sec \theta - \tan \theta = \frac{1}{x}$। এখন, $(\sec \theta + \tan \theta) - (\sec \theta - \tan \theta) = x - \frac{1}{x}$ বা, $2 \tan \theta = \frac{x^2-1}{x}$ বা, $\tan \theta = \frac{x^2-1}{2x}$। আবার, $(\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) = x + \frac{1}{x}$ বা, $2 \sec \theta = \frac{x^2+1}{x}$ বা, $\sec \theta = \frac{x^2+1}{2x}$। অতএব, $$ \begin{aligned} \sin \theta &= \frac{\tan \theta}{\sec \theta} \\ &= \frac{x^2-1}{2x} \times \frac{2x}{x^2+1} \\ &= \frac{x^2-1}{x^2+1} \end{aligned} $$