why $\tan x = \frac{\sin x}{\cos x}$? and not $\tan x$ = opposite/adjacent? we know that $\tan x =\left(\frac{\text{opposite}}{\text{adjacent}}\right)$, but sometimes I see that $\tan x = (\frac{\sin x}{\cos x})$, is that the same thing or why it is different sometimes?
cause when $\tan x =\left(\frac{\text{opposite}}{\text{adjacent}}\right)$:
$\tan x = \frac{1}{2}$ - for example
but sometimes is:
$\tan x = \frac{1}{5}$ or $ \frac{2}{5}$ - if hypotenuse is $= 5$
why is that?
 A: You may also want to search for trig explanations using the unit circle (here is an example)
and see the connections between the opposite leg (height $y$) and $\sin(\theta)$, and the adjacent leg (length $x$) and $\cos(\theta)$. You'll see immediately why your two versions of tangent are the same.
A: The relation that you write as
$$
\tan x=\frac{\text{opposite}}{\text{adjacent}}
$$
only holds for acute angles and was, of course, the original definition of the tangent, in times when just positive numbers were considered. Since, for acute angles we have
$$
\frac{\text{opposite}}{\text{adjacent}}=
\frac{\text{opposite}}{\text{hypothenuse}}\cdot
\frac{\text{hypothenuse}}{\text{adjacent}}=
\sin x\cdot \frac{1}{\cos x}=\frac{\sin x}{\cos x}
$$
this relation has been taken as the definition of the tangent function also for any angle $x$ such that $\cos x\ne0$.
So, while the “opposite/adjacent” definition is handy for doing computations on triangles, the $“\sin x/\cos x”$ definition is better for analytic computations involving trigonometric functions. They agree for acute angles $x$ and that's the important point to note.
A: Roughly, they are the same definition of tangent:
$$\begin{align} \require{cancel}
  \tan  = \frac{\sin}{\cos} 
       &= \frac{\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)}
              {\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)} \\
       &= \left(\frac{\text{opposite}}{\text{hypotenuse}}\right) \div
         \left(\frac{\text{adjacent}}{\text{hypotenuse}}\right) \\
       &= \left(\frac{\text{opposite}}{\cancel{\text{hypotenuse}}}\right) \times
         \left(\frac{\cancel{\text{hypotenuse}}}{\text{adjacent}}\right) \\
       &= \frac{\text{opposite}}{\text{adjacent}}
\end{align}$$
