What are the area of a triangle with side lengths $\tan(x)$, $\cos(x)$ and $\sin(x)$? 
Consider a non­degenerated right triangle with sides of length $\sin x$, $\cos x$, and $\tan x$ where $x$ is a real number.
  Compute the possible values of the area of this triangle.



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*I was thinking of more along the lines of Heron's formula but that was very nasty indeed, rather I have attempted to find out what are the lengths of the triangle and I have done the Pythagorean theorem which was in vain.  

*Another thing that I have done was to graph all three and to see which one was the largest in the y-value but it turned out that at times one graph was larger than the other and at other times it wasn't.

*And another thing that I have done was to plug and chug in values such as the number 3 into all of the trig functions and do Pythagorean theorem to see if it satisfied it but none of them didn't
The reason why that I have done those steps was so that I can multiply the legs and divide by two to find the area of the triangle but in order to do that I must know what are the legs.
I was wondering if there was any other way?

Since $\sin x$, $\cos x$ and $\tan x$ are all positive, we may assume that $\displaystyle x\in\left(0,\frac{\pi}{2}\right)$. Therefore, $\tan x>\sin x$.
If $\cos x>\tan x$, then $\cos x$ is the hypotenuse and 
\begin{align}
\sin^2x+\tan^2x&=\cos^2x\\
\sin^2x(\cos^2x+1)&=\cos^4x\\
1-\cos^4x&=\cos^4x\\
\cos x&=\frac{1}{\sqrt[4]{2}}
\end{align}
The area of the triangle is
\begin{align}
\frac{1}{2}\sin x\tan x&=\frac{\sin^2 x}{2\cos x}\\
&=\frac{1-\cos^2 x}{2\cos x}\\
&=\frac{\displaystyle1-\frac{1}{\sqrt{2}}}{\displaystyle\frac{2}{\sqrt[4]{2}}}\\
&=\frac{\sqrt[4]{2}(2-\sqrt{2})}{4}\\
\end{align}
If $\tan x>\cos x$, then $\tan x$ is the hypotenuse and 
\begin{align}
\sin^2x+\cos^2x&=\tan^2x\\
x&=\frac{\pi}{4}
\end{align}
The area of the triangle is
$$\frac{1}{2}\sin x\cos x=\frac{1}{4}$$
