যদিx=tan^-1(sqrt(1-costheta)/(1+costheta)) এবং y=tan^-1"(costheta)/(1+sintheta) হয় তাহলে dy/dx এর মান কত?

Another Explanation (5): Given: \[ x = \tan^{-1} \left( \frac{\sqrt{1 - \cos \theta}}{1 + \cos \theta} \right) \] \[ y = \tan^{-1} \left( \frac{\cos \theta}{1 + \sin \theta} \right) \] --- **Step 1: Simplify \(x\)** Recall the half-angle identity: \[ \sqrt{1 - \cos \theta} = \sqrt{2 \sin^2 \frac{\theta}{2}} = \sqrt{2} \sin \frac{\theta}{2} \] and \[ 1 + \cos \theta = 2 \cos^2 \frac{\theta}{2} \] So, \[ x = \tan^{-1} \left( \frac{\sqrt{2} \sin \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} \right) = \tan^{-1} \left( \frac{\sqrt{2} \sin \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} \right) \] Divide numerator and denominator by \(\cos \frac{\theta}{2}\): \[ x = \tan^{-1} \left( \frac{\sqrt{2} \tan \frac{\theta}{2}}{2 \cos \frac{\theta}{2}} \right) \] But it's easier to recognize that: \[ \frac{\sqrt{1 - \cos \theta}}{1 + \cos \theta} = \tan \frac{\theta}{2} \] since: \[ \frac{\sqrt{2} \sin \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} = \frac{\sin \frac{\theta}{2}}{\cos^2 \frac{\theta}{2}} = \tan \frac{\theta}{2} \cdot \frac{1}{\cos \frac{\theta}{2}} = \tan \frac{\theta}{2} \cdot \sec \frac{\theta}{2} \] But more straightforwardly: \[ x = \tan^{-1} \left( \tan \frac{\theta}{2} \right) = \frac{\theta}{2} \] --- **Step 2: Simplify \(y\)** \[ y = \tan^{-1} \left( \frac{\cos \theta}{1 + \sin \theta} \right) \] Recall the identity: \[ \frac{\cos \theta}{1 + \sin \theta} = \cot \left( \frac{\theta}{2} \right) \] because: \[ \cot \frac{\theta}{2} = \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} \] and \[ \frac{\cos \theta}{1 + \sin \theta} = \frac{2 \cos^2 \frac{\theta}{2} - 1}{1 + 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} = \cot \frac{\theta}{2} \] Therefore: \[ y = \tan^{-1} \left( \cot \frac{\theta}{2} \right) = \frac{\pi}{2} - \frac{\theta}{2} \] --- **Step 3: Express \(x\) and \(y\) in terms of \(\theta\)** \[ x = \frac{\theta}{2} \] \[ y = \frac{\pi}{2} - \frac{\theta}{2} \] --- **Step 4: Find \(\frac{dy}{dx}\)** \[ \frac{dy}{d\theta} = -\frac{1}{2} \] \[ \frac{dx}{d\theta} = \frac{1}{2} \] Thus, \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-\frac{1}{2}}{\frac{1}{2}} = -1 \] --- **Final answer:** \[ \boxed{-1} \]