Q: Prove derivative of an implicit function The question:
$\sqrt{1+x^2} + \sqrt{1+y^2} = a(x-y)$ prove that $\dfrac{dy}{dx} = \sqrt{\dfrac{1+y^2}{1+x^2}}$
I tried doing it the normal way and got
$\dfrac{dy}{dx} = \dfrac{(a\sqrt{1+x^2}-x)(\sqrt{1+y^2})}{(a\sqrt{1+y^2} + y)(\sqrt{1+x^2})}$
I also tried doing a trigonometry sub to get a relation of $a$ into a independent constant
$x=\tan(\alpha)$
$y=\tan(\beta)$
And got
$$a = \dfrac{\cos(\alpha) + \cos(\beta)}{\sin(\alpha - \beta)}$$ I don't know if this one can be further manipulated.
Is there something I missed? or is there another way of solving it?

A: Note that $$a=\frac{\sqrt{1+x^{2}}+\sqrt{1+y^{2}}}{x-y}.$$
so that
$$
\begin{align*}
\frac{a\sqrt{1+x^{2}}-x}{a\sqrt{1+y^2}+y} &=\frac{\frac{(1+x^{2})+\sqrt{1+x^{2}}\sqrt{1+y^{2}}-x^2+xy}{x-y}}{\frac{(1+y^{2})+\sqrt{1+x^{2}}\sqrt{1+y^2}+xy-y^{2}}{x-y}}\\
&=\frac{1+\sqrt{1+x^{2}}\sqrt{1+y^{2}}+xy}{1+\sqrt{1+x^{2}}\sqrt{1+y^{2}}+xy}\\
&=1.
\end{align*}$$
Therefore from your own calculations
$$\frac{dy}{dx}=\frac{(a\sqrt{1+x^{2}}-x)(\sqrt{1+y^{2}})}{(a\sqrt{1+y^{2}}+y)(\sqrt{1+x^{2}})}=\frac{\sqrt{1+y^{2}}}{\sqrt{1+x^{2}}}.$$

A: Need to solve it by using trig substituion and complex number
$$\sqrt{1+x^2} + \sqrt{1+y^2} = a(x-y)$$ can be written as
$$\sqrt{1-(ix)^2} + \sqrt{1-(iy)^2} = -ai(ix-iy)$$
here the function is defined as $0 \leq (ix)^2 \leq 1$ and so for $(iy)^2$
$$ix = \sin(\alpha) \tag{i}$$
$$iy = \sin(\beta)\tag{ii}$$
So the equation will be now
$$\cos(\alpha) + \cos(\beta) = -ai(\sin(\alpha) -\sin(\beta))$$
using $\cos(C) + \cos(D)$ and $\sin(C) - \sin(D)$
\begin{align*}
\cos(\dfrac{\alpha - \beta}{2})&= -ai(\sin(\dfrac{\alpha-\beta}{2}))\\
\dfrac{\alpha - \beta}{2} &= \text{arccot}(-ai)\\
\alpha - \beta &= 2\text{arccot}(-ai)
\end{align*}
from (i) & (ii)
$$\arcsin(ix) - \arcsin(iy) = 2\text{arccot}(-ai)$$
differentiating both sides
\begin{align*}\dfrac{i}{\sqrt{1+x^2}} - \dfrac{i\dfrac{dy}{dx}}{\sqrt{1+y^2}} &= 0\\
\dfrac{dy}{dx} &= \sqrt{\dfrac{1+y^2}{1+x^2}}
\end{align*}
