Question

If $\tan A,\tan B$ are the roots of the quadratic $ab{x^2} - {c^2}x + ab = 0$ , where a, b, c are the sides of triangle, then prove that $\cos C = 0$ .

Answer 1

Hint: In this question, we use the concept of quadratic equation. If quadratic equation is in form of $p{x^2} + qx + r = 0$ so the sum and product of roots are $\dfrac{{ - q}}{p}$ and $\dfrac{r}{p}$ respectively. We also use trigonometric identity $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ .

Complete step-by-step answer:
We have a quadratic equation $ab{x^2} - {c^2}x + ab = 0$ and $\tan A,\tan B$ are roots of this quadratic equation.
Now, we have to find the sum and product of roots of the above quadratic equation. If quadratic equation is in form of $p{x^2} + qx + r = 0$ so the sum and product of roots are $\dfrac{{ - q}}{p}$ and $\dfrac{r}{p}$ respectively .
$
  {\text{Sum of roots, }}\tan A + \tan B = \dfrac{{ - \left( { - {c^2}} \right)}}{{ab}} = \dfrac{{{c^2}}}{{ab}}...........\left( 1 \right) \\
  {\text{Product of roots, }}\tan A \times \tan B = \dfrac{{ab}}{{ab}} = 1..................\left( 2 \right) \\
$
Now, we use trigonometric identity $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ .
From (1) and (2) equation.
$
   \Rightarrow \tan \left( {A + B} \right) = \dfrac{{\dfrac{{{c^2}}}{{ab}}}}{{1 - 1}} \\
   \Rightarrow A + B = \dfrac{\pi }{2} \\
 $
We know A, B and C are angles of the triangle whose sides are a, b and c. So, the sum of all angles of the triangle is 1800.
$
   \Rightarrow A + B + C = \pi \\
   \Rightarrow \dfrac{\pi }{2} + C = \pi \\
   \Rightarrow C = \dfrac{\pi }{2} \\
$
Now, $\cos C = \cos \dfrac{\pi }{2} = 0$
So, it’s proved that $\cos C = 0$ .

Note: In such types of problems we use some important points. First we find the sum and product of roots and put value in trigonometric identity. Then apply the property of the triangle that the sum of all angles of the triangle is 1800. So, we will get the required answer.