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

Toolbox:
• To find $\large\frac{dy}{dx}$ for implict function $\large\frac{d}{dx}$$\phi(y)=\large\frac{d}{dy}$$\phi(y).\large\frac{dy}{dx}$
• A function $f(x,y)$ is said to be implicit if it is jumbled in such a way,that it is not possible to write $y$ exclusively s a function of $x$.
Step 1:
Given : $\sqrt {1-x^2}+\sqrt {1-y^2}=a(x-y),$
Put $x=\sin A$ and $y=\sin B$
$\Rightarrow A=\sin^{-1}x$ and $B=\sin^{-1}y$
$\sqrt{1-\sin ^2A}+\sqrt{1-\sin^2B}=a(\sin A-\sin B)$
But $1-\sin^2A=\cos^2A$
$1-\sin^2B=\cos^2B$
$\Rightarrow \cos A+\cos B=a(\sin A-\sin B)$----(1)
But $\cos A+\cos B=2\large\frac{\cos(A+B)}{2}.\frac{\cos(A-B)}{2}$
$\sin A-\sin B=2\large\frac{\sin(A-B)}{2}.\frac{\cos(A+B)}{2}$
Now substituting the above in equ(1) we get,
$2\cos\big(\large\frac{A+B}{2}\big).$$\cos\big(\large\frac{A-B}{2}\big)=a\bigg[$$2\sin\big(\large\frac{A-B}{2}\big)$$-\cos\big(\large\frac{A+B}{2}\big)\bigg] \Rightarrow \cot\big(\large\frac{A-B}{2}\big)$$=a$
$\Rightarrow \cot^{-1}a=\large\frac{A-B}{2}$
$\Rightarrow 2\cot^{-1}a={A-B}$
Step 2:
Substituting for $A$ and $B$ we get,
$2\cot^{-1}a=\sin^{-1}x-\sin^{-1}y$
Now differentiating w.r.t $x$ we get,
$0=\large\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}.\frac{dy}{dx}$
Therefore $\large\frac{dy}{dx}=\large\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$
Hence proved.