Area of triangle under tan conditions.

Question

For an acute triangle $ABC$, the following conditions hold.

$$\frac{1}{\tan A} + \frac{1}{\tan B} + \frac{1}{\tan C } =2$$

$$ a^2 + b^2 + c^2 =50 $$

Compute the area of such a triangle.

I used a tan addition law and get, $$\tan A + \tan B + \tan C = \tan A \tan B \tan C $$.

And let $ \tan A \tan B \tan C = k $, so $\tan A , \tan B , \tan C $ are roots of the following cubic equation. $x^3 - k x^2 +2k x - k $. I am stuck right here.

Answer 1

We have $$2=\sum_{cyc}\frac{ab\cos\gamma}{ab\sin\gamma}=\frac{\sum\limits_{cyc}(a^2+b^2-c^2)}{4S}=\frac{50}{4S},$$ which gives $$S=\frac{25}{4}.$$