MAO 1991

$\frac{tan^{2}x}{3}+\frac{1}{7}=\frac{-2tanx+1+tan^{2}x}{10}$

$=49tan^{2}x+42tanx+9=0$

$(7tanx+3)^{2}=0$

$tanx=- \frac{3}{7}$

Rearranging the two sides of the equation, noting that $\sin 2x = 2 \sin x \cos x$

$\displaystyle\frac{7\sin^2 x}{30}+\displaystyle\frac{\sin x \cos x}{5} +\displaystyle\frac{3\cos^2 x}{70} = 0$

$\Leftrightarrow \displaystyle\frac{7\sin^2 x}{30}+\displaystyle\frac{\sin x \cos x}{5} +\displaystyle\frac{\cos^2 x}{7} - \displaystyle\frac{1}{10} = 0$

As $\cos x = 0$ doesn't satisfy the equation, divide both sides by $\displaystyle\frac{\cos^2 x}{210}$ gives us:

$49 \tan^2 x + 42 \tan x + 9 = 0$ , or: $(7\tan x)^2 + 2\times 3 \times 7\tan x + 3^2 = 0$

$\Leftrightarrow (7\tan x + 3)^2 = 0$

$\Leftrightarrow \tan x = \displaystyle\frac{3}{7}$

So, $a+b=3+7=\boxed{10}$