Solveeit Logo

Question

Question: If \(\tan A+\tan B+\tan C=\tan A.\tan B.\tan C\) then: a). A, B, C must be the angle of the triang...

If tanA+tanB+tanC=tanA.tanB.tanC\tan A+\tan B+\tan C=\tan A.\tan B.\tan C then:
a). A, B, C must be the angle of the triangle
b). the sum of any two is equal to the third
c). A + B + C must be the integral multiple of π\pi
d). none of the above

Explanation

Solution

To solve the above question we will use the trigonometric formula, i.e. tan(A+B)=tanA+tanB1tanA.tanB\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B} after arranging the equation given in the question and we will also use the formula tanθ=tan(θ)-\tan \theta =\tan \left( -\theta \right) and the solution of trigonometric equation tanθ=tanα\tan \theta =\tan \alpha , that is θ=nπ+α\theta =n\pi +\alpha

Complete step-by-step solution
It is given in the question that:
tanA+tanB+tanC=tanA.tanB.tanC\tan A+\tan B+\tan C=\tan A.\tan B.\tan C
So, we can rewrite the above equation as:
tanA+tanB=tanA.tanB.tanCtanC\Rightarrow \tan A+\tan B=\tan A.\tan B.\tan C-\tan C
Taking tanC-\tan C common from the above equation we will get:
tanA+tanB=tanC(tanA.tanB+1)\Rightarrow \tan A+\tan B=-\tan C\left( -\tan A.\tan B+1 \right)
tanC=tanA+tanB(tanA.tanB+1)\Rightarrow -\tan C=\dfrac{\tan A+\tan B}{\left( -\tan A.\tan B+1 \right)}
tanC=tanA+tanB(1tanA.tanB)(1)\Rightarrow -\tan C=\dfrac{\tan A+\tan B}{\left( 1-\tan A.\tan B \right)}----(1)
Now, we know that tan(A+B)=tanA+tanB1tanA.tanB\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B}.
So, we will put the value of tan(A+B)\tan \left( A+B \right) in the above equation (1), then we will get:
tanC=tan(A+B)-\tan C=\tan \left( A+B \right)
tan(A+B)=tanC\Rightarrow \tan \left( A+B \right)=-\tan C
Now, we know that tanθ=tan(θ)-\tan \theta =\tan \left( -\theta \right), so we can write the above equation as:
tan(A+B)=tan(C)\Rightarrow \tan \left( A+B \right)=\tan \left( -C \right)
Now, from standard solution of trigonometric equation we know that if tanθ=tanα\tan \theta =\tan \alpha , then we can say that θ=nπ+α\theta =n\pi +\alpha .
So, for the trigonometric equation tan(A+B)=tan(C)\Rightarrow \tan \left( A+B \right)=\tan \left( -C \right), we can say that:
(A+B)=nπ+(C)\left( A+B \right)=n\pi +\left( -C \right)
Hence, A+B+C=nπA+B+C=n\pi
Hence, we can say that A + B + C must be the integral multiple of π\pi .
So, option (c) is the correct option.

Note: Students are required to have memorized the trigonometric formulas and standard solution of trigonometric equations otherwise they will not be able to solve the above question. And, also there are so many chances of making mistakes in this question because we directly remember thattanA+tanB+tanC=tanA.tanB.tanC\tan A+\tan B+\tan C=\tan A.\tan B.\tan C, where A, B, and C are the angle of the triangle and we mark option (a) as correct but it is wrong this time because in this option it is saying must but after solving we are getting A + B + C as an integral multiple of π\pi and also if the question is the single choice we always have to choose the best possible answer so option (c) is correct.