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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...

saladi
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    Related : http://math.stackexchange.com/questions/1849467/given-abc-180-circ-find-value-of-tan-a-cdot-tan-b-tan-b-cdot-tan-c-t/1849508#1849508 Observe that " the product of the sines of the angles " is not required :) – lab bhattacharjee Jul 15 '16 at 13:32
  • tanA , tanB , tanC are the roots of a cubic ....................yes the angles A , B , C are the same as the angles in the triangle that I took sines and cosines of. – saladi Jul 15 '16 at 13:35
  • Thanks.. I got the answer . But please do tell me if there is any shorter way of doing such problems .... – saladi Jul 15 '16 at 13:48
  • Welcome! Sorry don't any better solution. – lab bhattacharjee Jul 15 '16 at 13:51

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