Compute the least possible area of a non-degenerate right triangle with sides of lengths $\sin{x}$, $\cos{x}$ and $\tan{x}$ where $x$ is a real number.
Let $\alpha$ be the included angle of sides $\sin x$ and $\cos x$. Then $$\cos\alpha=\frac{\sin^2 x + \cos^2 x - \tan^2 x}{2\sin x\cos x}=\frac{1-\tan^2 x}{2\sin x\cos x}$$ Therefore $$\sin\alpha=\frac{\sqrt{(2\sin x\cos x)^2-(1-\tan^2 x)^2}}{2\sin x\cos x}$$
It follows that the area of this triangle equals $$\frac{1}{2}\sin{x}\cos{x}\frac{\sqrt{(2\sin x\cos x)^2-(1-\tan^2 x)^2}}{2\sin x\cos x}=\frac{1}{4}\sqrt{(2\sin x\cos x)^2-(1-\tan^2 x)^2}$$
The part inside the square root operator can be simplied to a function respect to $\cos^2x$ where $0< \cos^2x \le 1$. Minimize that function can lead to the final answer as: $$\boxed{\frac{\sqrt[4]{2}}{2}-\frac{\sqrt[4]{8}}{4}}$$