Question

If \[y = x\tan \dfrac{{\alpha + \beta }}{2}\], then find \[\tan A + \tan B + \tan C\]
A. \[\dfrac{{a + b + c}}{{abc}}\]
B. 0
C. \[\tan A\tan B\tan C\]
D. \[\tan A\tan B + \tan B\tan C + \tan C\tan A\]

Answer 1

Hint: We will use the formula of the tangent sum of two angles. Then simplify the equation using the sum of all angles of a triangle and the formula of supplementary angle in trigonometry.

Formula used:
Tangent sum of two angles
\[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]
Supplementary angle in trigonometry
\[\tan \left( {\pi - \theta } \right) = - \tan \theta \]

Complete step by step solution:
The formula of tangent sum of two angles is
\[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] ….(i)
We know that the sum of all angles of a triangle is \[\pi \].
Thus, \[A + B + C = \pi \]
\[ \Rightarrow A + B = \pi - C\]
Substitute \[A + B = \pi - C\] in equation (i)
\[ \Rightarrow \tan \left( {\pi - C} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]
Apply the formula of supplementary angle in trigonometry:
\[ \Rightarrow - \tan C = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]
Cross multiply:
\[ \Rightarrow - \tan C\left( {1 - \tan A\tan B} \right) = \tan A + \tan B\]
\[ \Rightarrow - \tan C + \tan A\tan B\tan C = \tan A + \tan B\]
\[ \Rightarrow + \tan A\tan B\tan C = \tan A + \tan B + \tan C\]
This is an identity and it is applicable for each triangle that is why the given relation between x and y is irrelevant here.
Hence option C is the correct option.

Additional information:
This formula \[\tan A\tan B\tan C = \tan A + \tan B + \tan C\] can be applied to all types of triangles. Here A, B, and C are the angles of the triangle.

Note: We can solve the given question using any two angles of the triangle. We can use \[\tan \left( {B + C} \right)\] and \[\tan \left( {A + C} \right)\] to get the result.