(a)OR(b)

If $\displaystyle \sqrt{1-\mathrm{x}^{2}}+\sqrt{1-\mathrm{y}^{2}}=\mathrm{a}(\mathrm{x}-\mathrm{y})$, prove that $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{\frac{1-\mathrm{y}^{2}}{1-\mathrm{x}^{2}}}$.

$$\begin{aligned} & \sqrt{1-\mathrm{x}^{2}}+\sqrt{1-\mathrm{y}^{2}}=\mathrm{a}(\mathrm{x}-\mathrm{y}) \\ & \text { Put } \mathrm{x}=\sin \theta, \mathrm{y}=\sin \phi \\ & \Rightarrow \cos \theta+\cos \phi=\mathrm{a}(\sin \theta-\sin \phi) \\ & \Rightarrow 2 \cos \left(\frac{\theta+\phi}{2}\right) \cos \left(\frac{\theta-\phi}{2}\right)=2 \mathrm{a} \sin \left(\frac{\theta-\phi}{2}\right) \cos \left(\frac{\theta+\phi}{2}\right) \\ & \Rightarrow \cot \left(\frac{\theta-\phi}{2}\right)=\mathrm{a} \\ & \Rightarrow \theta-\phi=2 \cot ^{-1} \mathrm{a} \\ & \Rightarrow \sin ^{-1} \mathrm{x}-\sin ^{-1} \mathrm{y}=2 \cot ^{-1} \mathrm{a} \\ & \Rightarrow \frac{1}{\sqrt{1-\mathrm{x}^{2}}}-\frac{1}{\sqrt{1-\mathrm{y}^{2}}} \frac{d \mathrm{y}}{d \mathrm{x}}=0 \\ & \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=\sqrt{\frac{1-\mathrm{y}^{2}}{1-\mathrm{x}^{2}}} \end{aligned} \begin{aligned} & \mathrm{y}=(\tan \mathrm{x})^{\mathrm{x}} \\ & \log \mathrm{y}=\mathrm{x} \log (\tan \mathrm{x}) \\ & \frac{1}{\mathrm{y}} \frac{d \mathrm{y}}{d \mathrm{x}}=\mathrm{x}\left(\frac{\sec ^{2} \mathrm{x}}{\tan \mathrm{x}}\right)+\log (\tan \mathrm{x}) \\ & \frac{d \mathrm{y}}{d \mathrm{x}}=(\tan \mathrm{x})^{\mathrm{x}}\left[\left(\frac{\mathrm{x} \sec ^{2} \mathrm{x}}{\tan \mathrm{x}}\right)+\log (\tan \mathrm{x})\right] \end{aligned} $$