# If sqrt(1 - x^2) + sqrt(1 - y^2) = a(x - y). Then how will you prove that dy/dx = sqrt[(1 - y^2)/(1 - x^2)]??

Jul 24, 2017

Refer to the Explanation.

#### Explanation:

Observe that, $\sqrt{1 - {x}^{2}} \mathmr{and} \sqrt{1 - {y}^{2}}$ are Meaningful, iff,

$| x | \le 1 , \mathmr{and} , | y | \le 1. \ldots \ldots \ldots \ldots . . \left({\star}^{1}\right) .$

This means that, there is no Harm if we let,

$x = \sin \theta , \mathmr{and} , y = \sin \phi \ldots \ldots \ldots \ldots \ldots \ldots . . \left({\star}^{2}\right) .$

$\therefore \sqrt{1 - {x}^{2}} + \sqrt{1 - {y}^{2}} = a \left(x - y\right) , \text{ becomes, }$

$\cos \theta + \cos \phi = a \left(\sin \theta - \sin \phi\right) .$

$\therefore 2 \cos \left(\frac{\theta + \phi}{2}\right) \cos \left(\frac{\theta - \phi}{2}\right) = 2 a \left\{\cos \left(\frac{\theta + \phi}{2}\right) \sin \left(\frac{\theta - \phi}{2}\right)\right\} .$

$\therefore \cos \left(\frac{\theta - \phi}{2}\right) = a \sin \left(\frac{\theta - \phi}{2}\right) .$

$\cot \left(\frac{\theta - \phi}{2}\right) = a .$

$\left(\frac{\theta - \phi}{2}\right) = a r c \cot a .$

$\therefore \theta - \phi = 2 a r c \cot a .$

From, $\left({\star}^{1}\right) , \mathmr{and} , \left({\star}^{2}\right) ,$ we have,

$a r c \sin x - a r c \sin y = 2 a r c \cot a .$

$\therefore \frac{d}{\mathrm{dx}} \left\{a r c \sin x - a r c \sin y\right\} = \frac{d}{\mathrm{dx}} \left\{2 a r c \cot a\right\} .$

$\therefore \frac{1}{\sqrt{1 - {x}^{2}}} - \frac{1}{\sqrt{1 - {y}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left\{y\right\} = 0 , \ldots \text{[The Chain Rule].}$

$\therefore \frac{1}{\sqrt{1 - {x}^{2}}} - \frac{1}{\sqrt{1 - {y}^{2}}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0.$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{1 - {y}^{2}}}{\sqrt{1 - {x}^{2}}} .$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \sqrt{\frac{1 - {y}^{2}}{1 - {x}^{2}}} .$

Enjoy Maths.!