If sqrt {1 - {x^2}} + sqrt {1 - {y^2}} = a(x | vedclass

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(B) Let $x = \sin 	heta$ and $y = \sin \phi$. Substituting these into the given equation: $\cos 	heta + \cos \phi = a(\sin 	heta - \sin \phi)$ Usin

Vedclass · Accepted Answer

$\sqrt {\frac{1 - {y^2}}{1 - {x^2}}} $

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(B) Let $x = \sin 	heta$ and $y = \sin \phi$. Substituting these into the given equation: $\cos 	heta + \cos \phi = a(\sin 	heta - \sin \phi)$ Using the sum-to-product formulas: $2 \cos \left( \frac{	heta + \phi}{2} ight) \cos \left( \frac{	heta - \phi}{2} ight) = a \left[ 2 \cos \left( \frac{	heta + \phi}{2} ight) \sin \left( \frac{	heta - \phi}{2} ight) ight]$ Assuming $\cos \left( \frac{	heta + \phi}{2} ight) 
eq 0$,we divide both sides: $\cot \left( \frac{	heta - \phi}{2} ight) = a$ $\frac{	heta - \phi}{2} = \cot^{-1} a \Rightarrow 	heta - \phi = 2 \cot^{-1} a$ Substituting back $	heta = \sin^{-1} x$ and $\phi = \sin^{-1} y$: $\sin^{-1} x - \sin^{-1} y = 2 \cot^{-1} a$ Differentiating both sides with respect to $x$: $\frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = 0$ $\frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} = \sqrt{\frac{1 - y^2}{1 - x^2}}$.

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

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