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Question: If the sum of the angles A, B and C is given by \(A+B+C={{180}^{\circ }}\) Then prove that \(\...

If the sum of the angles A, B and C is given by
A+B+C=180∘A+B+C={{180}^{\circ }}
Then prove that
tan⁡(A2)tan⁡(B2)+tan⁡(B2)tan⁡(C2)+tan⁡(C2)tan⁡(A2)=1\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

Explanation

Solution

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⁡C2\tan \dfrac{C}{2}common from the last two terms, express the angle C2\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=πA+B+C=\pi
Therefore,
A2+B2=π2−C2=90∘−C2..............(1.1)\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⁡(A2)tan⁡(B2)+tan⁡(B2)tan⁡(C2)+tan⁡(C2)tan⁡(A2)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⁡(C2)\tan \left( \dfrac{C}{2} \right) common from the last two terms, we obtain
LHS=tan⁡(A2)tan⁡(B2)+tan⁡(C2)(tan⁡(B2)+tan⁡(A2))........(1.2)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
tan⁡(a+b)=tan⁡(a)+tan⁡(b)1−tan⁡(a)tan⁡(b) ⇒tan⁡(a)+tan⁡(b)=tan⁡(a+b)(1−tan⁡(a)tan⁡(b)).......(1.4) \begin{aligned} & \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{aligned}
Using equation (1.4) in equation (1.2), by taking a=A and b=B, we get
LHS=tan⁡(A2)tan⁡(B2)+tan⁡(C2)tan⁡(A+B2)(1−tan⁡(A2)tan⁡(B2))........(1.5)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∘{{90}^{\circ }} minus some angle should be equal to cotangent of the angle. Therefore,
tan⁡(90∘−θ)=cot⁡(θ)=1tan⁡(θ)........(1.6)\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⁡(A+B2)=tan⁡(90∘−C2)=1tan⁡(C2)........(1.7)\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
LHS=tan⁡(A2)tan⁡(B2)+tan⁡(C2)1tan⁡(C2)(1−tan⁡(A2)tan⁡(B2)) =tan⁡(A2)tan⁡(B2)+1−tan⁡(A2)tan⁡(B2) =1........................(1.8) \begin{aligned} & 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{aligned}
Thus, from equation (1.8), we have proved that
LHS=tan⁡(A2)tan⁡(B2)+tan⁡(B2)tan⁡(C2)+tan⁡(C2)tan⁡(A2)=1=RHSLHS=\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⁡(B2)\tan \left( \dfrac{B}{2} \right) common from the first two terms and then used equation (1.4) for the angles A2\dfrac{A}{2} and C2\dfrac{C}{2}. However, the final result would still remain the same.