Algebra: solve for $x$ from a trigonometric expression $\frac{tan25}{tan20}=\frac{\left(90+x\right)}{40+x}$
$x =\frac{90\tan \left(20^{\circ \:}\right)-40\tan \left(25^{\circ \:}\right)}{\tan \left(25^{\circ \:}\right)-\tan \left(20^{\circ \:}\right)}$
$x=137.82851$
Now instead of cross multiplying and then solve for x, I did this:
$\frac{tan25}{tan20}=\frac{90+x-\left(40+x\right)+\left(40+x\right)}{40+x}$
$=\frac{50}{40+x} + 1$
Some algrebra and it gives you:
$\frac{50tan\left(20\right)}{tan\left(25\right)}-90$
$x =-50.97316$
How come I get 2 different answers from 2 different methods? What did I did wrong? What is the difference?
 A: The first method is correct. But since you haven't given the steps that you took in the second method, I cannot know exactly where you went wrong. But I have a pretty good guess:
You started with
$$
\frac{\tan 25}{\tan 20} = \frac{90+x}{40+x} = \frac{50}{40+x}+1
$$
So far, this is correct. Now the goal is to take $()^{-1}$ on both sides, in order to get the $x$ to the numerator instead of denominator. You probably did something like this: First move the $+1$ ...
$$\tag{2}
\frac{\tan 25}{\tan 20} -1  = \frac{50}{40+x} \qquad \Rightarrow \qquad \frac{\tan 20}{\tan 25} -1  = \frac{40+x}{50}
$$
This would lead to the second result that you gave. But do you notice the (quite drastic) error that was made? The inverse of a sum is not the sum of inverses (as on the left-hand side). So the logic arrow doesn't hold in Equation (2) and it is incorrect. Do you understand why this is incorrect?
So basically the inverse
$$
\left(\frac{\tan 25}{\tan 20} -1  \right)^{-1} \neq \left( \frac{\tan 25}{\tan 20}\right)^{-1} - 1^{-1}
$$
Instead, it should be calculated like this:
$$
\left(\frac{\tan 25}{\tan 20} -1  \right)^{-1} = \left(\frac{\tan 25}{\tan 20} -\frac{\tan 20}{\tan 20}  \right)^{-1} = \left(\frac{\tan 25 - \tan 20}{\tan 20} \right)^{-1} = \frac{\tan 20}{\tan 25 - \tan 20}
$$
