Question

If the sum of the angles A, B and C is given by
$A+B+C={{180}^{\circ }}$
Then prove that
$\tan \left( \dfrac{A}{2} \right)\tan \left( \dfrac{B}{2} \right)+\tan \left( \dfrac{B}{2} \right)\tan \left( \dfrac{C}{2} \right)+\tan \left( \dfrac{C}{2} \right)\tan \left( \dfrac{A}{2} \right)=1$

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. Also, we can $\tan \dfrac{C}{2}$common from the last two terms, express the angle $\dfrac{C}{2}$ in terms of the angle A and B, and then use the formula for tangent of 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,
$\dfrac{A}{2}+\dfrac{B}{2}=\dfrac{\pi }{2}-\dfrac{C}{2}={{90}^{\circ }}-\dfrac{C}{2}..............(1.1)$
The left hand side of the equation is
$LHS=\tan \left( \dfrac{A}{2} \right)\tan \left( \dfrac{B}{2} \right)+\tan \left( \dfrac{B}{2} \right)\tan \left( \dfrac{C}{2} \right)+\tan \left( \dfrac{C}{2} \right)\tan \left( \dfrac{A}{2} \right)$
Taking $\tan \left( \dfrac{C}{2} \right)$ common from the last two terms, we obtain
$LHS=\tan \left( \dfrac{A}{2} \right)\tan \left( \dfrac{B}{2} \right)+\tan \left( \dfrac{C}{2} \right)\left( \tan \left( \dfrac{B}{2} \right)+\tan \left( \dfrac{A}{2} \right) \right)........(1.2)$
Now, the formula for tangent of sum of two angles is given by
$\begin{align}
  & \tan \left( a+b \right)=\dfrac{\tan (a)+\tan (b)}{1-\tan (a)\tan (b)} \\
 & \Rightarrow \tan (a)+\tan (b)=\tan \left( a+b \right)\left( 1-\tan (a)\tan (b) \right).......(1.4) \\
\end{align}$
Using equation (1.4) in equation (1.2), by taking a=A and b=B, we get
$LHS=\tan \left( \dfrac{A}{2} \right)\tan \left( \dfrac{B}{2} \right)+\tan \left( \dfrac{C}{2} \right)\tan \left( \dfrac{A+B}{2} \right)\left( 1-\tan \left( \dfrac{A}{2} \right)\tan \left( \dfrac{B}{2} \right) \right)........(1.5)$
Now, the tangent of ${{90}^{\circ }}$ minus some angle should be equal to cotangent of the angle. Therefore,
$\tan \left( {{90}^{\circ }}-\theta \right)=\cot \left( \theta \right)=\dfrac{1}{\tan \left( \theta \right)}........(1.6)$
Now, using equation (1.1) in equation (1.6), we get
$\tan \left( \dfrac{A+B}{2} \right)=\tan \left( {{90}^{\circ }}-\dfrac{C}{2} \right)=\dfrac{1}{\tan \left( \dfrac{C}{2} \right)}........(1.7)$
Using equation (1.7), we obtain
$\begin{align}
  & LHS=\tan \left( \dfrac{A}{2} \right)\tan \left( \dfrac{B}{2} \right)+\tan \left( \dfrac{C}{2} \right)\dfrac{1}{\tan \left( \dfrac{C}{2} \right)}\left( 1-\tan \left( \dfrac{A}{2} \right)\tan \left( \dfrac{B}{2} \right) \right) \\
 & =\tan \left( \dfrac{A}{2} \right)\tan \left( \dfrac{B}{2} \right)+1-\tan \left( \dfrac{A}{2} \right)\tan \left( \dfrac{B}{2} \right) \\
 & =1........................(1.8) \\
\end{align}$
Thus, from equation (1.8), we have proved that
$LHS=\tan \left( \dfrac{A}{2} \right)\tan \left( \dfrac{B}{2} \right)+\tan \left( \dfrac{B}{2} \right)\tan \left( \dfrac{C}{2} \right)+\tan \left( \dfrac{C}{2} \right)\tan \left( \dfrac{A}{2} \right)=1=RHS$
Which was asked in the question.

Note: In this question, we could also have taken $\tan \left( \dfrac{B}{2} \right)$ common from the first two terms and then used equation (1.4) for the angles $\dfrac{A}{2}$ and $\dfrac{C}{2}$. However, the final result would still remain the same.