Question

If $A + B + C = \pi ,n \in {\rm Z}$, then find $\tan nA + \tan nB + \tan nC$.
A) $0$
B) $1$
C) $\tan nA\tan nB\tan nC$
D) None of these

Answer 1

Hint: Using the sum formula of $\tan $ we can solve the question. Also remember $\tan $ has zero values for integer multiples of $\pi $. Here also it is given that \[n\] belongs to the set of Integers. So combining all these we have we can find the answer of the question.

Formula used:
For any integer $n,\tan n\pi = 0$
For any $A,B,C$, we have,
$\tan (A + B + C) = \dfrac{{\tan A + \tan B + \tan C - \tan A\tan B\tan C}}{{1 - \tan A\tan B - \tan B\tan C - \tan A\tan C}}$

Complete step-by-step answer:
In the question it is given that
$A + B + C = \pi $ and $n \in {\rm Z}$ (Set of integers)
Multiplying both sides of the above equation by $n$ we get,
$ \Rightarrow n(A + B + C) = nA + nB + nC = n\pi $
We can apply $\tan $ on both sides.
$ \Rightarrow \tan (nA + nB + nC) = \tan n\pi $
We know $\tan $ takes value zero on integer multiples of $\pi $.
So, we have $\tan n\pi = 0,n \in {\rm Z}$.
$ \Rightarrow \tan (nA + nB + nC) = 0 - - - (i)$
For any $A,B,C$, we have,
$\tan (A + B + C) = \dfrac{{\tan A + \tan B + \tan C - \tan A\tan B\tan C}}{{1 - \tan A\tan B - \tan B\tan C - \tan A\tan C}}$
Using this we get,
$\tan (nA + nB + nC) = \dfrac{{\tan nA + \tan nB + \tan nC - \tan nA\tan nB\tan nC}}{{1 - \tan nA\tan nB - \tan nB\tan nC - \tan nA\tan nC}}$
Substituting the above expression in equation $(i)$, we have,
$\dfrac{{\tan nA + \tan nB + \tan nC - \tan nA\tan nB\tan nC}}{{1 - \tan nA\tan nB - \tan nB\tan nC - \tan nA\tan nC}} = 0$
This means the numerator of the above equation is zero.
$ \Rightarrow \tan nA + \tan nB + \tan nC = \tan nA\tan nB\tan nC$
Therefore, the answer is option C.

Additional information:
If there were only $A$ and $B$, then $\tan A + B = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
The equation of $A,B$ and $C$ is actually derived from this one by combining two at a time.
There is also another formula for the case of subtraction.
$\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
This formula too can be extended to the sum of three variables.

Note:Since $\tan$ takes the value zero on integer multiples of $\pi $, we could reduce the expression to get the answer. For $\sin $ and $\cos $ it is not the case. They do not take value zero on every integer multiple of $\pi $. So while doing trigonometric operations we should be aware and careful about the values taken by the functions.