Problem on maximum area Problem

A sector subtends an angle $2\alpha$ at the centre then the greatest area of the rectangle inscribed in it would be:

I have two solutions for this:
Solution 1:

Solution 2:

Which one is incorrect and why?
 A: In the first "solution", you cannot write $\cos \alpha = AB/BC$ because $ABC$ isn't a right triangle [In fact it is the tangent in point $A$ that makes a  $90°$ angle with radius $PA$].
Your second solution is perfect: you have chosen the good parameter $\theta$ to express it, and I see no single shorter route to the solution.
This said, I propose you a solution in the way people like Fermat or Huyghens were able to obtain solutions to optimality issues before the discovery of calculus by Leibnitz and Newton. Take a look at the figure below, where we have taken $R=1$ with abscissas of $A,B$ being computed as $\sin \theta \cot \alpha$ and $\cos \theta$ by elementary trigonometry as in the second solution of the OP.

Let us consider the consequences of an infinitesimal change of $\theta$ into $\theta+d\theta$ with (half) rectangle $ABCD$ changed into $A'B'C'D'$.
Area changes can be measured in terms of area gain (green rectangle) vs. area loss (the two red rectangles). Please note that, as all is done at infinitesimal order $d\theta$, the areas of yellow triangles (enlarged above the principal figure) are neglected, being in $d\theta^2$.
The lengths of these rectangles are easily computed ; their widths are obtained by easy computations on yellow triangles .
The key remark is that the optimum is reached when gains and losses are balanced:
$$\underbrace{AB \times D'E}_{\color{green}{\text{gain}}}= \underbrace{AD \times DE+BC \times FC}_{\color{red}{\text{losses}}}$$
$$\underbrace{(\cos \theta - \sin \theta \cot \alpha) \cos \theta d\theta}_{\color{green}{\text{gain}}}= \underbrace{\sin \theta (\cos \theta \cot \alpha + \sin \theta)d\theta}_{\color{red}{\text{losses}}}$$
$$ \cos^2 \theta - \sin^2 \theta=2 \sin \theta \cos \theta \cot \alpha$$
$$ \cos 2 \theta=\sin 2 \theta \cot \alpha$$
We find back:
$$\cot 2 \theta=\cot \alpha  \ \implies \theta = \frac12 \alpha$$
(because we are working in $(0, \frac{\pi}{2})$).
