Question

If$A+B+C=\pi $, then using trigonometric identities, prove that
$\tan A+\tan B+\tan C=\tan A\tan B\tan C$

Answer 1

Hint: In this question, the sum of the angles is given and we have to find the relation between the tangents of these angles. Therefore, using the information, we can express the sum of two angles in terms of the other angle and $\pi $ and then use the formula for the tangent of the sum of angles to obtain the expression given in the question.

Complete step-by-step answer:
In the question, the sum of the three angles A, B and C is given to be
$A+B+C=\pi $
Therefore,
$A+B=\pi -C..............(1.1)$
Taking the tangent on both sides we get,
$\tan \left( A+B \right)=\tan \left( \pi -C \right)..............(1.2)$
Now, we the formula for tangent of sum of two angles is given by
$\tan \left( a+b \right)=\dfrac{\tan (a)+\tan (b)}{1-\tan (a)\tan (b)}.......(1.2)$
And the tangent of two angles is given by
$\tan \left( a-b \right)=\dfrac{\tan (a)-\tan (b)}{1+\tan (a)\tan (b)}.......(1.3)$
Therefore, using equations (1.2) and (1.3) in equation (1.2), we obtain
$\dfrac{\tan (A)+\tan (B)}{1-\tan (A)\tan (B)}=\dfrac{\tan (\pi )-\tan (C)}{1+\tan (\pi )\tan (C)}.......(1.4)$

Now, as the angle $\pi $ is equal to ${{180}^{\circ }}$, therefore its tangent is given by
$\tan (\pi )=\dfrac{\sin (\pi )}{\cos (\pi )}=\dfrac{0}{-1}=0$
Therefore using the value of $\tan (\pi )$ in equation (1.4), we obtain
$\dfrac{\tan (A)+\tan (B)}{1-\tan (A)\tan (B)}=\dfrac{0-\tan (C)}{1+0\times \tan (C)}=-\tan (C).......(1.5)$
Cross multiplying the numerator and denominator, we obtain
$\begin{align}
  & \tan (A)+\tan (B)=-\tan (C)\left( 1-\tan (A)\tan (B) \right)=-\tan (C)+\tan (A)\tan (B)\tan (C) \\
 & \Rightarrow \tan (A)+\tan (B)+\tan (C)=\tan (A)\tan (B)\tan (C) \\
\end{align}$
Which is exactly the equation that we wanted to prove. Thus, we have successfully proved that
$\tan (A)+\tan (B)+\tan (C)=\tan (A)\tan (B)\tan (C)$

Note: To prove the equation, we used the given relation that the sum of the angles is ${{180}^{\circ }}$. Now, the sum of the angles of a triangle is also ${{180}^{\circ }}$. Therefore, the same method will also be applicable to prove the relation for the angles of a triangle. Also, in equation (1.2), we could have also used the relation that
$\tan (\pi -\theta )=-\tan \left( \theta \right)$ to simplify the RHS directly without using equation (1.3).