Question

If A+B+C= \[{{180}^{\circ }}\] , and tanA+tanB+tanC=k(\[tanA\cdot tanB\cdot tanC\]), then value of k is

Answer 1

Hint:
As the sum of all the three angles is \[{{180}^{\circ }}\] , hence it can be inferred from this fact that these are the angles of a triangle as the sum of all the three angles of a triangle is also \[{{180}^{\circ }}\] .
Another important formula that is used in the solution is the formula for finding the tangent or the tan value of the sum of two angles which is as follows
\[\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\cdot \tan B}\]

Complete answer:
As mentioned in the question, we know that these are the angles of a triangle as the sum of all the three angles of a triangle is also \[{{180}^{\circ }}\] .
Now, on using the formula for finding the tangent or the tan value of the sum of two angles as mentioned in the hint, we get
\[\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\cdot \tan B}\ \ \ \ \ ...(a)\]
Now, here we can see that the sum of the two angles can be written as
A+B= \[{{180}^{\circ }}\] - C …(b)
Now, we can use equation (a) and (b) to get
\[\tan ({{180}^{\circ }}-C)=\dfrac{\tan A+\tan B}{1-\tan A\cdot \tan B}\ \ \ \ \ ...(c)\]
On using the fact that
\[\tan ({{180}^{\circ }}-C)=\tan C\]
So, on cross multiplying the equation (c) and using the above mentioned fact, we get
\[\begin{align}
  & \tan C=\dfrac{\tan A+\tan B}{1-\tan A\cdot \tan B}\ \\
 & \tan C-\tan A\cdot \tan B\cdot \tan C=\tan A+\tan B \\
 & \tan A\cdot \tan B\cdot \tan C=\tan A+\tan B+\tan C\ \ \ \ \ ...(d) \\
\end{align}\]
Now, on comparing equation (d) and the question, we get that value of k is 1.

Note:
In this question, if the students don’t figure out that these are the angles of a triangle as the sum of all the three angles of a triangle is also \[{{180}^{\circ }}\] then the question can become a little tricky and the students can make an error and then they would not get the correct answer.
Also, the property of the tan that is used is also very crucial.