Prove that $ (a+b) \cos(x) = a-b$, if $ \sin(x) + \tan(x) = \frac{4a\sqrt{ab}}{a^2-b^2},$ $ \tan(x) - \sin(x) = \frac{4a\sqrt{ab}}{a^2-b^2}.$ Prove that $ (a+b) \cos(x) = a-b$, if $$ \sin(x) + \tan(x) = \frac{4a\sqrt{ab}}{a^2-b^2},$$
$$ \tan(x) - \sin(x) = \frac{4a\sqrt{ab}}{a^2-b^2}.$$
I tried solving it with system, but with no result.

grand_chat has already pointed out that you may want 
$$\tan x-\sin x=\frac{4\color{red}{b}\sqrt{ab}}{a^2-b^2}\tag 1$$
With 
$$\tan x+\sin x=\frac{4a\sqrt{ab}}{a^2-b^2}\tag 2$$
now $(1)+(2)$ gives
$$2\tan x=\frac{4(a+b)\sqrt{ab}}{(a+b)(a-b)}\Rightarrow \tan x=\frac{2\sqrt{ab}}{a-b}$$
Also, $(2)-(1)$ gives
$$2\sin x=\frac{4(a-b)\sqrt{ab}}{(a-b)(a+b)}\Rightarrow \sin x=\frac{2\sqrt{ab}}{a+b}$$
Here, use
$$\tan x=\frac{\sin x}{\cos x}\Rightarrow \cos x=\frac{\sin x}{\tan x}=\left(\frac{2\sqrt{ab}}{a+b}\right)/\left(\frac{2\sqrt{ab}}{a-b}\right)=\frac{a-b}{a+b}.$$
