Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

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


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

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.

share|cite|improve this question
Use the half angle formulas from to transform everything into polynomial equalities in two variables. – Mariano Suárez-Alvarez Nov 3 '10 at 6:12
Why haven't you accepted an answer to this question? – anonymous Mar 18 '11 at 16:43
up vote 7 down vote accepted

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 $$\tan\Bigl(\frac{C}{2}\Bigr) \cdot \Bigl[ \tan\Bigl(\frac{A}{2}\Bigr) + \tan\Bigl(\frac{B}{2}\Bigr) \Bigr] = 1 \cdot \Bigl[ 1 - \tan\Bigl(\frac{A}{2}\Bigr) \cdot \tan\Bigl(\frac{B}{2}\Bigr)\Bigr]$$

share|cite|improve this answer
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 3 '10 at 10:46
@Debanjan: Edited! before you wrote that comment – anonymous Nov 3 '10 at 10:47
Both showing 1 mint ago so I guess simultaneous posting,however I will +1 for a clear approach. – Quixotic Nov 3 '10 at 10:48

More generally, we have from De Moivre's theorem

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


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

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 $n=3.$

share|cite|improve this answer



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.


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

Act II.


Simplify the above expression.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.