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

Answer: 10

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