Question

Let the product of the sines of the angles of a 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.

Answer 1

Given $\sin A \sin B \sin C =\frac{2}{3} ; \cos A \cos B \cos C =\frac{1}{9} ; \tan A \tan B \tan C =\frac{2}{3} \cdot \frac{9}{1}=6$
$\therefore \tan A +\tan B +\tan C =6 \left[ As\right.$, in $\left.\triangle ABC , \sum \tan A =\prod \tan A \right]$
Hence the cubic is $x^3-6 x^2+\left(\sum \tan A \tan B\right) x^2-6=0$....(1)
Now
$\sum \tan A \tan B=\frac{\sin A \sin B \cos C+\sin B \sin C \cos A+\sin C \sin A \cos B}{\cos A \cos B \cos C}$ ....(2)
Now
$A+B+C=\pi $
$\cos (A+B+C)=-1 $
$\cos (A+B) \cos C-\sin (A+B) \sin C=-1 $
$(\cos A \cos B-\sin A \sin B) \cos C-\sin C(\sin A \cos B+\cos A \sin B)=-1 $
$1+\cos A \cos B \cos C=\sin A \sin B \cos C+\sin B \sin C \cos A+\sin C \sin A \cos B$ .....(3)
substituting in (2), we get
$\sum \tan A \tan B=\frac{1+\cos A \cos B \cos C}{\cos A \cos B \cos C}=\frac{1+\frac{1}{9}}{\frac{1}{9}}=\frac{10}{9} \cdot \frac{9}{1}=10$
Hence the cubic is $x^3-6 x^2+10 x-6=0$
Clearly, the sum of the products of the roots taken two at a time is 10.