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प्रश्न
If \[\sqrt{1 - x^2} + \sqrt{1 - y^2} = a \left( x - y \right)\] , prove that \[\frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{1 - x^2}\] ?
उत्तर
\[\text{ We have }, \sqrt{1 - x^2} + \sqrt{1 - y^2} = a\left( x - y \right)\]
\[\text{Let x } = \sin A , y = \sin B\]
\[ \Rightarrow \sqrt{1 - \sin^2 A} + \sqrt{1 - \sin^2 B} = a\left( \sin A - \sin B \right)\]
\[ \Rightarrow \cos A + \cos B = a\left( \sin A - \sin B \right) \]
\[ \Rightarrow a = \frac{\cos A + \cos B}{\sin A - \sin B}\]
\[ \Rightarrow a = \frac{2 \cos\frac{A + B}{2}\cos\frac{A - B}{2}}{2 \cos\frac{A + B}{2}\sin\frac{A - B}{2}} ...........[\because \sin A - \sin B = 2 \cos\frac{A + B}{2}\sin\frac{A - B}{2} \text{ and } \cos A + \cos B = 2 \cos\frac{A + B}{2}\cos\frac{A - B}{2}]\]
\[ \Rightarrow a = \cot\left( \frac{A - B}{2} \right)\]
\[ \Rightarrow \cot^{- 1} a = \frac{A - B}{2}\]
\[ \Rightarrow 2 \cot^{- 1} a = A - B\]
\[ \Rightarrow 2 \cot^{- 1} a = \sin^{- 1} x - \sin^{- 1} y ..........\left[ \because x = \sin A, y = \sin B \right]\]
Differentiating with respect to x, we get,
\[\frac{d}{dx}\left( 2co t^{- 1} a \right) = \frac{d}{dx}\left( \sin^{- 1} x \right) - \frac{d}{dx}\left( \sin^{- 1} y \right)\]
\[ \Rightarrow 0 = \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}}\frac{d y}{d x}\]
\[ \Rightarrow \frac{1}{\sqrt{1 - y^2}}\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^2}}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}\]
\[ \Rightarrow \frac{d y}{d x} = \sqrt{\frac{1 - y^2}{1 - x^2}}\]
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