(a)OR(b)

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

Let $\displaystyle x=\sin A, \mathrm{y}=\sin B \Rightarrow A=\sin ^{-1} x, B=\sin ^{-1} \mathrm{y}$ \[\begin{aligned} & \therefore \sqrt{1-x^{2}}+\sqrt{1-\mathrm{y}^{2}}=\mathrm{a}(x-\mathrm{y}) \\ & \Rightarrow \cos A+\cos B=\mathrm{a}(\sin A-\sin B) \\ & \Rightarrow 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)=2 \mathrm{a} \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \\ & \Rightarrow \cot \left(\frac{A-B}{2}\right)=\mathrm{a} \Rightarrow A-B=2 \cot ^{-1} \mathrm{a} \\ & \Rightarrow \sin ^{-1} x-\sin ^{-1} \mathrm{y}=2 \cot ^{-1} \mathrm{a} \\ & \operatorname{differentiatebothsideswrt} x \\ & \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-\mathrm{y}^{2}}} \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=0 \\ & \Rightarrow \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=\sqrt{\frac{1-\mathrm{y}^{2}}{1-x^{2}}} \end{aligned} \] \[\begin{aligned} & x=\mathrm{a}\left(\cos \theta+\log \tan \frac{\theta}{2}\right) \\ & \Rightarrow \frac{\mathrm{d} x}{\mathrm{d} \theta}=\mathrm{a}\left(-\sin \theta+\frac{1}{\tan \dfrac{\theta}{2}} \times \sec ^{2} \frac{\theta}{2} \times \frac{1}{2}\right) \\ & \quad=\mathrm{a}\left(-\sin \theta+\frac{1}{\sin \theta}\right)=\mathrm{a}\left(\frac{1-\sin ^{2} \theta}{\sin \theta}\right) \\ & \frac{\mathrm{d} x}{\mathrm{d} \theta}=\mathrm{a} \cot \theta \cos \theta \\ & \text { Also, } \mathrm{y}=\sin \theta \Rightarrow \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \theta}=\cos \theta \\ & \therefore \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=\frac{\tan \theta}{\mathrm{a}} \end{aligned} \] Differetiating wrt $\displaystyle x$, \[\begin{aligned} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}} & =\frac{\sec ^{2} \theta}{\mathrm{a}} \times \frac{\mathrm{d} \theta}{\mathrm{d} x} \\ & =\frac{\sec ^{3} \theta \tan \theta}{\mathrm{a}^{2}} \end{aligned} \] $\displaystyle \left.\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}\right]_{\mathrm{at} \theta=\frac{\pi}{4}}=\frac{2 \sqrt{2}}{\mathrm{a}^{2}}$