Proving Legendres Relation for elliptic curves The legendre's relation can be stated as follows
$$
    K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*) = \frac{\pi}{2}
$$
where $k^* = \sqrt{1 - k^2}$ is the complimentary modulus, and $E$ and $K$ are
respectively complete elliptic integral's of the first and second kind
$$
K(k) = \int_0^{\pi/2} \frac{\mathrm{d}\theta}{\sqrt{1-k^2 \sin^2\theta}}\ ,\quad
E(k) = \int_0^{\pi/2}\sqrt {1-k^2 \sin^2\theta}\ \mathrm{d}\theta\,.
$$
Now sorry for not attempting to solve this problem myself, but I have tried both 
myself and finding sources online. Alas it seems this relation has been somewhat forgotten. I did however find an article claiming to show the relation, but I do not have access to check it's validity.
Can someone provide sources for a proof of this relation, or outline a proof? Hopefully not using hypergeometric functions, but a proof more in the spirit of Legendre.
Any help would be greatly appreciated. 
 A: Here is an outline of a proof:
First use the integral definitions to show that $$\frac{\mathrm d E(k)}{\mathrm dk} = \frac{E(k)-K(k)}{k} \tag{1}$$
and $$ \frac{\mathrm d K(k)}{\mathrm d k} = \frac{E(k) - (k^{*})^{2} K(k)}{k(k^{*})^{2}} \tag{2}.$$
Next use those identities to show that $$ \frac{\mathrm d}{\mathrm dk} \Big( K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*)  \Big) =0 .$$
This will show that $K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*) $ is a constant.
Finally take the limit as $k$ goes to $0$ from the right to show that the constant is $ \frac{\pi}{2}$.
EDIT:
The second identity is not obvious. See the question here.
SECOND EDIT:
I'm going to fill in the gaps a bit.
If we differentiate $E(k^{*})$ or $K(k^{*})$ with respect to $k^{*}$, we'll get similar formulas to the ones above since all we're really doing is relabeling.
Using the chain rule, we get $$ \begin{align} \frac{\mathrm d E(k^{*})}{ \mathrm dk} &= \frac{\mathrm d E(k^{*})}{\mathrm dk^{*}} \frac{\mathrm d k^{*}}{\mathrm dk} =  \frac{E(k^{*})-K(k^{*})}{k^{*}} \frac{-k}{\sqrt{1-k^{2}}} \\ &= \frac{E(k^{*})-K(k^{*})}{k^{*}} \frac{-k}{k^{*}} = -\frac{k}{(k^{*})^{2}} \Big( E(k^{*})-K(k^{*}) \Big) , \end{align}$$
and
$$ \begin{align} \frac{\mathrm d K(k^{*})}{\mathrm dk} &= \frac{\mathrm d K(k^{*})}{\mathrm dk^{*}} \frac{\mathrm d k^{*}}{\mathrm dk} = \frac{E(k^{*}) - k^{2} K(k^{*})}{k^{*} k^{2}} \frac{-k}{k^{*}} \\ &= - \frac{1}{k(k^{*})^{2}} \left( E(k^{*}) - k^{2} K(k^{*}) \right). \end{align}$$
Therefore,
$$ \begin{align} &\frac{\mathrm d}{\mathrm dk} \Big( K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*)  \Big) \\ &= \frac{E(k) - (k^{*})^{2}K(k)}{k (k^{*})^{2}}E(k^{*}) - K(k) \frac{k}{(k^{*})^{2}} \Big( E(k^{*})-K(k^{*}) \Big) \\ &+ \frac{E(k)-K(k)}{k} K(k^{*}) - E(k) \frac{1}{k(k^{*})^{2}} \left( E(k^{*}) - k^{2} K(k^{*}) \right) \\ &- \frac{E(k) - (k^{*})^{2}K(k)}{k (k^{*})^{2}} K(k^{*}) + K(k) \frac{1}{k(k^{*})^{2}} \left( E(k^{*})-k^{2}K(k^{*}) \right) \\ &=  E(k) E(k^{*}) \left( \frac{1}{k (k^{*})^{2}} - \frac{1}{k (k^{*})^{2}} \right) + K(k) E(k^{*}) \left(- \frac{1}{k} - \frac{k}{(k^{*})^{2}} + \frac{1}{k(k^{*})^{2}}\right) \\ &+ K(k) K(k^{*}) \left( \frac{k}{(k^{*})^{2}} - \frac{1}{k} + \frac{1}{k} - \frac{k}{(k^{*})^{2}}\right) + E(k) K(k^{*}) \left( \frac{1}{k} + \frac{k}{(k^{*})^{2}} - \frac{1}{k (k^{*})^{2}} \right) \\ &= K(k) E(k^{*}) \left(- \frac{1}{k} - \frac{k}{(k^{*})^{2}} + \frac{1}{k(k^{*})^{2}}\right)  - E(k) K(k^{*}) \left( -\frac{1}{k} - \frac{k}{(k^{*})^{2}} + \frac{1}{k (k^{*})^{2}} \right) . \end{align}$$
But since
$$ - \frac{1}{k} - \frac{k}{(k^{*})^{2}} + \frac{1}{k(k^{*})^{2}} = \frac{-(k^{*})^{2} - k^{2} +1}{k(k^{*})^{2}} = \frac{-(1-k^{2})-k^{2}+1}{k(k^{*})^{2}} = 0,$$
we have
$$ \frac{\mathrm d}{\mathrm dk} \Big( K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*)  \Big) =0. $$
And
$$ \begin{align} \lim_{k \to 0^{+}} \Big( K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*) \Big) &= \frac{\pi}{2} (1) + \lim_{k \to 0^{+}}  \Big( E(k) -K(k) \Big) K(k^{*})  \\ & \overset{(3)}{=} \frac{\pi}{2} + \lim_{k \to 0^{+}} \mathcal{O}(k^{2}) \left(-\ln (k) + \mathcal{O}(1) \right) \\ &= \frac{\pi}{2} + 0 \\ &= \frac{\pi}{2}. \end{align}$$

$(3)$ The answer here shows that $K(k) = -\log(k^{*})+ \mathcal{O}(1) $ as $k \to 1^{-}$.  Replace $k$ with $k^{*}$.
