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Question: If \(A + B + C = \pi ,n \in {\rm Z}\), then find \(\tan nA + \tan nB + \tan nC\). A) \(0\) B) \(...

If A+B+C=π,n∈ZA + B + C = \pi ,n \in {\rm Z}, then find tan⁡nA+tan⁡nB+tan⁡nC\tan nA + \tan nB + \tan nC.
A) 00
B) 11
C) tan⁡nAtan⁡nBtan⁡nC\tan nA\tan nB\tan nC
D) None of these

Explanation

Solution

Using the sum formula of tan⁡\tan we can solve the question. Also remember tan⁡\tan has zero values for integer multiples of π\pi . Here also it is given that nn 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⁡nπ=0n,\tan n\pi = 0
For any A,B,CA,B,C, we have,
tan⁡(A+B+C)=tan⁡A+tan⁡B+tan⁡C−tan⁡Atan⁡Btan⁡C1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Atan⁡C\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=πA + B + C = \pi and n∈Zn \in {\rm Z} (Set of integers)
Multiplying both sides of the above equation by nn we get,
⇒n(A+B+C)=nA+nB+nC=nπ\Rightarrow n(A + B + C) = nA + nB + nC = n\pi
We can apply tan⁡\tan on both sides.
⇒tan⁡(nA+nB+nC)=tan⁡nπ\Rightarrow \tan (nA + nB + nC) = \tan n\pi
We know tan⁡\tan takes value zero on integer multiples of π\pi .
So, we have tan⁡nπ=0,n∈Z\tan n\pi = 0,n \in {\rm Z}.
⇒tan⁡(nA+nB+nC)=0−−−(i)\Rightarrow \tan (nA + nB + nC) = 0 - - - (i)
For any A,B,CA,B,C, we have,
tan⁡(A+B+C)=tan⁡A+tan⁡B+tan⁡C−tan⁡Atan⁡Btan⁡C1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Atan⁡C\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)=tan⁡nA+tan⁡nB+tan⁡nC−tan⁡nAtan⁡nBtan⁡nC1−tan⁡nAtan⁡nB−tan⁡nBtan⁡nC−tan⁡nAtan⁡nC\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)(i), we have,
tan⁡nA+tan⁡nB+tan⁡nC−tan⁡nAtan⁡nBtan⁡nC1−tan⁡nAtan⁡nB−tan⁡nBtan⁡nC−tan⁡nAtan⁡nC=0\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.
⇒tan⁡nA+tan⁡nB+tan⁡nC=tan⁡nAtan⁡nBtan⁡nC\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 AA and BB, then tan⁡A+B=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan A + B = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}
The equation of A,BA,B and CC 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)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\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⁡\tan takes the value zero on integer multiples of π\pi , we could reduce the expression to get the answer. For sin⁡\sin and cos⁡\cos it is not the case. They do not take value zero on every integer multiple of π\pi . So while doing trigonometric operations we should be aware and careful about the values taken by the functions.