Question

In a triangle $\vartriangle ABC$, $\tan A + \tan B + \tan C = 6$ and $\tan A\tan B = 2$, then the values of $\tan A,\tan B,\tan C$ are
1) 1,2,3
2) $3,\dfrac{2}{3},\dfrac{7}{3}$
3) $4,\dfrac{1}{2},\dfrac{3}{2}$
4) none of these

Answer 1

Hint: We will use the angle sum property of a triangle and will write the condition, $\angle A + \angle B + \angle C = {180^ \circ }$ and then we will take tan on both sides. Then, apply the formula $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$, to simplify the expression. We will cross multiply and substitute the given values to get the required values.

Complete step-by-step answer:
We are given that $\tan A + \tan B + \tan C = 6$ and $\tan A\tan B = 2$.
We have to find the values of $\tan A,\tan B,\tan C$.
If $A,B$ and $C$ are three angles of a triangle, the sum of three angles is ${180^ \circ }$.
Then, $\angle A + \angle B + \angle C = {180^ \circ }$
And $\angle A + \angle B = {180^ \circ } - \angle C$
Taking tan on both sides.
$\tan \left( {\angle A + \angle B} \right) = \tan \left( {{{180}^ \circ } - \angle C} \right)$
We know that $\tan \left( {{{180}^ \circ } - \theta } \right)$ lies in second quadrant and is equal to $ - \tan \theta $
$\tan \left( {\angle A + \angle B} \right) = - \tan \angle C$
Now, we will apply the formula, $\tan \left( {\angle A + \angle B} \right)$ which is $\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Then,
$\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = - \tan C$
On cross-multiplying the equation, we will get,
$
  \tan A + \tan B = - \tan C + \tan A\tan B\tan C \\
   \Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C \\
$
Now, we will substitute the values of $\tan A + \tan B + \tan C = 6$ and $\tan A\tan B = 2$ in the above equation
$
  6 = 2\tan C \\
  \tan C = 3 \\
$
We will substitute this value in $\tan A + \tan B + \tan C = 6$
$\tan A + \tan B + 3 = 6$
$ \Rightarrow \tan A + \tan B = 3$ eqn. (1)
Now, we know that ${\left( {a + b} \right)^2} - 4ab = {\left( {a - b} \right)^2}$
Then,
${\left( {\tan A + \tan B} \right)^2} - 4\tan A\tan B = {\left( {\tan A - \tan B} \right)^2}$
Substitute the known values in the
$
  {\left( 3 \right)^2} - 4\left( 2 \right) = {\left( {\tan A - \tan B} \right)^2} \\
   \Rightarrow 9 - 8 = {\left( {\tan A - \tan B} \right)^2} \\
$
$ \Rightarrow \tan A - \tan B = 1$ eqn. (2)
We will add equation (1) and equation (2) to find the value of $\tan A$
$
  \tan A + \tan B + \tan A - \tan B = 1 + 3 \\
   \Rightarrow 2\tan A = 4 \\
   \Rightarrow \tan A = 2 \\
$
Substitute the value of \[\tan A\] in equation (1) to find the value of $\tan B$.
$
  2 + \tan B = 3 \\
   \Rightarrow \tan B = 1 \\
$
Hence, the value of $\tan A$ is 2, $\tan B$ is 1 and $\tan C$ is 3.
Hence, option A is correct.

Note: Since, $A,B$ and $C$ are three angles of a triangle, their sum is equal to ${180^ \circ }$. One must the formula of $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ to do this question correctly, Similar type of question is also possible with cot ratio of triangle.