Given a line in the first quadrant of the cartesian plane tangent to the unit circle, consider the triangle formed by this line and the positive $x$ and $y$ axis. Prove that the minimal area of this triangle is $1$, using derivation.
My approach was to say $y = mx + b$ (where $m < 0$ and $b > 1$). Above is the minimal case, where $m = -1, b = \sqrt{2}$ to give an area of $1$. My intuition is that rotating any tangent line will yield that the "sum of the intercepts" on the $x$ and $y$ axis will increase, yielding a larger area of the triangle, and hence the above case would be the minimal. Maybe realting the monotonity of $\tan(x)$ on the interval $(0, \frac{\pi}{2}$) to the length of the tangent and then using Pythagoras to get the lenghts of the base and height of the triangle? However, I couldn't make this rigorous and more importantly, couldn't relate any of this to derivatives, since I couldn't find any area function only in terms of one variable ($A(m)$ or $A(k)$)) and then finding the minimum using a derivative. Then, I realized that the part of the unit circle in the first quadrant is given by $y = \sqrt{1-x^2}$, and hence one could set $kx + m = \sqrt{1-x^2}$ and $\frac{d}{dx} [kx + m] = \frac{d}{dx} [\sqrt{1-x^2}]$ and try to make make some progress from there. I, however, got a really complicated algebraic expression, and considering that this question was from a no calculator and no digital tools test, I would like to find a simpler and more elegant way...
Consider the x and y intercepts for the line as $b$ and $h$ respectively, the equation of the line is $$ \frac{x}{b} + \frac{y}{a} = 1$$
The area of the triangle is $A = \tfrac{1}{2} b h$, but first we need to enforce the tangency condition.
Looking at the trig circle geometry we have that
$$ \begin{aligned} b & = \frac{1}{\cos \theta} & h & = \frac{1}{\sin \theta} \end{aligned} $$
This makes the area of the triangle
$$ A = \frac{1}{2 \sin \theta \cos \theta} = \frac{1}{ \sin( 2 \theta ) }$$
The smallest area possible is when $\sin(2 \theta)=1$ which corresponds to the $\theta = \tfrac{\pi}{4}$ orientation of the line.
You can also arrive at this result from a symmetry viewpoint. Note that switching x and y should not change the geometry of the problem, so the triangle remains unchanged under a $y=x$ mirror operation. This is only possible when $\theta = \tfrac{\pi}{4}$.