প্রদত্ত, $$\begin{aligned} x &= \tan^{-1} \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} \\ &= \tan^{-1} \sqrt{\frac{2\sin^2\frac{ heta}{2}}{2\cos^2\frac{ heta}{2}}} \\ &= \tan^{-1} \left( \tan\frac{ heta}{2} \right) \\ &= \frac{ heta}{2} \end{aligned}$$ এবং $$\begin{aligned} y &= \tan^{-1} \left( \frac{\cos\theta}{1+\sin\theta} \right) \\ &= \tan^{-1} \left( \frac{\sin(\frac{\pi}{2}-\theta)}{1+\cos(\frac{\pi}{2}-\theta)} \right) \\ &= \tan^{-1} \left( \frac{2\sin(\frac{\pi}{4}-\frac{\theta}{2})\cos(\frac{\pi}{4}-\frac{\theta}{2})}{2\cos^2(\frac{\pi}{4}-\frac{\theta}{2})} \right) \\ &= \tan^{-1} \left( \tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right) \right) \\ &= \frac{\pi}{4} - \frac{ heta}{2} \end{aligned}$$ যেহেতু $x = \frac{\theta}{2}$, তাই $y = \frac{\pi}{4} - x$। অতএব, $$\begin{aligned} \frac{dy}{dx} &= \frac{d}{dx} \left( \frac{\pi}{4} - x \right) \\ &= 0 - 1 \\ &= -1 \end{aligned}$$