Question

If $p$ is the product of the sines of angles of a triangle and $q$ the product of their cosines, the tangents of the angle are roots of the equation

• A. $q{x}^3-p{x}^2+(1+q)x-p=0$
• B. $p{x}^3-q{x}^2+(1+p)x-q=0$
• C. $(1+q)x^3-p{x}^2+(1+q)x-p=0$
• D. none of the above

Answer 1

Answer: (A)

$q{x}^3-p{x}^2+(1+q)x-p=0$

Here $p=\sin A\sin B\sin C$ and $q=\cos A\cos B\cos C$
$\therefore \frac{p}{q}=\tan A\tan B\tan C$ and
$\tan A\tan B+\tan B\tan C+\tan C\tan A$
$=\frac{\sin A}{\cos A}\frac{\sin B}{\cos B}+\frac{\sin B}{\cos B}\frac{\sin C}{\cos C}+\frac{\sin C}{\cos C}\frac{\sin A}{\cos A}$
$=\frac{\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B}{\cos A\cos B\cos C}$
$=\frac{\sin B(\sin A\cos C+\cos A\sin C)+\sin C\sin A\cos B}{\cos A\cos B\cos C}$
$=\frac{\sin B\sin (A+C)+\sin C\sin A\cos B}{\cos A\cos B\cos C}$
$=\frac{\sin B\sin (\pi -B)+\sin C\sin A\cos B}{\cos A\cos B\cos C}$
$=\frac{{\sin }^{2}B+\sin C\sin A\cos B}{\cos A\cos B\cos C}$
$=\frac{1-{\cos }^{2}B+\sin C\sin A\cos B}{\cos A\cos B\cos C}$
$=\frac{1-\cos B(\cos B-\sin A\sin C)}{\cos A\cos B\cos C}$
$=\frac{1-\cos B(\cos B-\sin A\sin C)}{q}$
$=\frac{1-\cos B(-\cos (A+C)-\sin A\sin C)}{q}$
$=\frac{1+\cos B(\cos A\cos C+\sin A\sin C-\sin A\sin C)}{q}$
$=\frac{1+\cos A\cos B\cos C}{q}$
$\frac{1+q}{q}$
and $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ in a $△ABC$
$=\frac{p}{q}$
$\therefore$Required equation is
$x^3-\left(\frac{p}{q}\right)x^2+\left(\frac{1+q}{q}\right)x-\frac{p}{q}=0$
or $q{x}^3-p{x}^2+(1+q)x-p=0$