<s>[INST] If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$? The question states:

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

The options are:

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

I tried differentiating the entire expression:
$$\begin{align}
\frac{-2x}{2\sqrt{1-x^2}}+\frac{-2y}{2\sqrt{1-y^2}}\frac{dy}{dx} &= a(1-\frac{dy}{dx}) \\
a + \frac{x}{\sqrt{1-x^2}} &= \frac{dy}{dx}(a-\frac{y}{\sqrt{1-y^2}}) \\
\frac{dy}{dx} &= (a+\frac{x}{\sqrt{1-x^2}})(a-\frac{y}{\sqrt{1-y^2}})^{-1}
\end{align}$$
I am not sure how to proceed now. Is there a better approach? Is it advisable to use my approach?
 [/INST] 
As stated in the comments this is actually a wee-bit lengthier and
  hence not recommended.

Since i always try to resort to trigonometric substitutions whenever fractional powers are involved I'm gonna attempt this by substituting $x=sinA$ and $y=sinB$
$cosA +cosB=a(sinA-sinB)$
$2cos\frac{(A+B)}{2} \times cos\frac{(A-B)}{2} =a\times 2Cos\frac{(A+B)}{2}\times Sin\frac{(A-B)}{2}$
$cot\frac{(A-B)}{2} =a$
$A-B=2cot^{-1}a$
Magically reducing it to 
$sin^{-1}x-sin^{-1}y =2cot^{-1}a$
Now differentiate…
$\frac{1}{\sqrt{1-x^{2}}} - \frac{1}{\sqrt{1-y^{2}}} \times \frac{dy}{dx}=0$
$\frac{dy}{dx}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$

If you are preparing for JEE ADV i would recommend practising differential calculus book by cengage , and for theory i would recommend Differential Calculus by AmitMAgarwal

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