Question

The product of the sine of the angles of a triangle is p and the product of their cosines is q. then prove that the tangents of the angles are the roots of the equation $q{x}^3-p{x}^2+(1+q)x-p=0$

Answer 1

From the question, $\sin A\sin B\sin C=p$ and $\cos A\cos B\cos C=q$
$\therefore$ $\tan A\tan B\tan C=p/q$ (i)
Also, $\tan A+\tan B+\tan C=\tan A\tan B\tan C=p/q$ (ii)
Now, $\tan A\tan B+\tan B\tan C+\tan C\tan A$
$=\frac{\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B}{\cos A\cos B\cos C}$
$=\frac{1}{2q}[({\sin }^{2}A+{\sin }^{2}B-{\sin }^{2}C)+({\sin }^{2}B+{\sin }^{2}C-{\sin }^{2}A)+({\sin }^{2}C+{\sin }^{2}A-{\sin }^{2}B)]$ [$\because A+B+C=\pi$and $2\sin A\sin B\sin C={\sin }^{2}A+{\sin }^{2}B-{\sin }^{2}C$]
$=\frac{1}{2q}[{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C]=\frac{1}{4q}[3-\left(\cos 2A\cos 2B\cos 2C\right)]$
$=\frac{1}{q}[1+\cos A\cos B\cos C]=\frac{1}{q}(1+q)$
The equation whose roots are $\tan A,\tan B,\tan C$ will be given by
$x^2-\left(\tan A\tan B\tan C\right)x^2+(\tan A\tan B+\tan B\tan C+\tan C\tan A)x-\tan A\tan B\tan C=0$
or $x^3-\frac{p}{q}{x}^2+\frac{1+q}{q}x-\frac{p}{q}=0$ or $q{x}^3-p{x}^2+(1+q)x-p=0$
Ans: 7