Let the product of the sines of the angles of the triangle is $\frac{2}{3}$ and the product of their cosines is $\frac{1}{9}.$ If $\tan A$ , $\tan B$ and $\tan C$ are the roots of the cubic, find the sum of the products of the roots taken two at a time.
My Approach: Since it is given product sines and cosines of a triangle we can calculate product of the roots the cubic by dividing . It is coming as $\frac{2}{27}$, and after that by using property of $\tan (A+B+C)$ we get sum and product of roots as equal. After this I tried to expand $\tan A \tan B + \tan B \tan C + \tan C \tan A$ in terms of sine and cosine but it is going very lengthy. Please suggest any appropriate method to solve this...