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The following two identities comes from my trigonometry module without any sort of proof,

If $A + B + C = \pi $ then,

$$\tan A + \tan B + \tan C = tan A \cdot tan B \cdot tan C$$

and,

$$ \tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{B}{2} \cdot \tan \frac{C}{2} + \tan \frac{C}{2} \cdot \tan \frac{A}{2} = 1 $$

PS:I am not much sure about whether the first one is fully correct or not, so if not please suggest the correct one and also I will be grateful if somebody suggest a suitable method (may be using Mathematica) to verify an identity like this prior to proving.

Jean Marie
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Quixotic
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  • Use the half angle formulas from http://en.wikipedia.org/wiki/Tangent_half-angle_formula#Identities to transform everything into polynomial equalities in two variables. – Mariano Suárez-Álvarez Nov 03 '10 at 06:12
  • Possible duplicate of https://math.stackexchange.com/questions/477364/tan-a-tan-b-tan-c-tan-a-tan-b-tan-c-in-a-triangle?noredirect=1 – Somos Mar 09 '24 at 00:15

3 Answers3

8

If $A+B+C= \pi \Longrightarrow \tan(A+B) = \tan(\pi -C) =-\tan(C)$. So we have $$\tan(A+B)= \frac{\tan(A) + \tan(B)}{1 - \tan(A)\cdot \tan(B)} = -\tan(C) $$ $$\Longrightarrow \tan(A)+\tan(B) = -\tan(C) \cdot \Bigl[1 - \tan(A)\tan(B)\Bigr]$$ from which the first one follows.

And for the second one, we have $\displaystyle\frac{A}{2} + \frac{B}{2} =\frac{\pi}{2}- \frac{C}{2} \Longrightarrow \tan\Bigl(\frac{A+B}{2}\Bigr)= \cot\Bigl(\frac{C}{2}\Bigr)$ Now expanding we have $$\tan\Bigl(\frac{A+B}{2}\Bigr)= \frac{\tan\Bigl(\frac{A}{2}\Bigr) + \tan\Bigl(\frac{B}{2}\Bigr)}{1- \tan\Bigl(\frac{A}{2}\Bigr)\cdot \tan\Bigl(\frac{B}{2}\Bigr)} = \cot\Bigl(\frac{C}{2}\Bigr)$$ Multiplying both sides by $\tan\frac{C}{2}$ we have $$\tan\Bigl(\frac{C}{2}\Bigr) \cdot \Bigl[ \tan\Bigl(\frac{A}{2}\Bigr) + \tan\Bigl(\frac{B}{2}\Bigr) \Bigr] = 1 \cdot \Bigl[ 1 - \tan\Bigl(\frac{A}{2}\Bigr) \cdot \tan\Bigl(\frac{B}{2}\Bigr)\Bigr]$$

  • In the second one you probably meant $\frac{A}{2} + \frac{B}{2} = ( \frac{\pi}{2} - \frac{C}{2} ) \Longrightarrow \tan\Bigl(\frac{A+B}{2}\Bigr)= \cot\Bigl(\frac{C}{2}\Bigr) $ – Quixotic Nov 03 '10 at 10:46
  • @Debanjan: Edited! before you wrote that comment –  Nov 03 '10 at 10:47
  • Both showing 1 mint ago so I guess simultaneous posting,however I will +1 for a clear approach. – Quixotic Nov 03 '10 at 10:48
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More generally, we have from De Moivre's theorem

$$\cos(\alpha_1+\alpha_2+\cdots+\alpha_n)= \text{Re}\prod_{k=1}^n(\cos \alpha_k + i \sin \alpha_k)$$

and

$$\sin(\alpha_1+\alpha_2+\cdots+\alpha_n)= \text{Im}\prod_{k=1}^n(\cos \alpha_k + i \sin \alpha_k)$$

and so

$$\tan(\alpha_1+\alpha_2+\cdots+\alpha_n)= \frac{ \text{Im}\prod_{k=1}^n ( 1 + i \tan \alpha_k) }{ \text{Re}\prod_{k=1}^n(1 + i \tan\alpha_k) }.$$

Consider the case $n=3.$

Derek Jennings
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1

Hints:

Prelude.

a. If $\alpha+\beta+\gamma=\pi$, then $\gamma=\pi-\alpha-\beta$.

b. $\tan(\pi-\theta)=-\tan(\theta)$

c. $\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$

d. $\tan\left(\frac{\pi}{2}-\theta\right)=\frac1{\tan(\theta)}$

Act I.

$$\tan(\alpha)+\tan(\beta)-\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}=-\tan(\alpha)\tan(\beta)\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$$

Can the left-hand side be made to look like the right-hand side?

Act II.

$$\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)+\left(\tan\left(\frac{\alpha}{2}\right)+\tan\left(\frac{\beta}{2}\right)\right)\left(\frac{\tan\left(\frac{\alpha}{2}\right)+\tan\left(\frac{\beta}{2}\right)}{1-\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)}\right)^{-1}$$

Simplify the above expression.

J. M. ain't a mathematician
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