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.

As written, the solution to the system of equations is $\sin x = 0$. I think you want 
$$\tan x -\sin x={4b\sqrt{ab}\over a^2-b^2}.$$ Then solving the system of equations yields
$$\sin x =\frac12 \left({4a\sqrt{ab}\over a^2-b^2}-{4b\sqrt{ab}\over a^2-b^2}\right)={2\sqrt{ab}\over a+b},$$
which gives $$\cos^2 x = 1 -\sin^2 x = {(a-b)^2\over(a+b)^2}.$$
