# Thread: Tangets of half angles in a triangle

1. ## Tangets of half angles in a triangle

Prove that in a triangle with angles $\displaystyle \alpha$, $\displaystyle \beta$ and $\displaystyle \gamma$:
$\displaystyle \tan\frac{\alpha}2\tan\frac{\beta}2+\tan\frac{\bet a}2\tan\frac{\gamma}2+\tan\frac{\gamma}2\tan\frac{ \alpha}2=1$.

2. Originally Posted by james_bond
Prove that in a triangle with angles $\displaystyle \alpha$, $\displaystyle \beta$ and $\displaystyle \gamma$:
$\displaystyle \tan\frac{\alpha}2\tan\frac{\beta}2+\tan\frac{\bet a}2\tan\frac{\gamma}2+\tan\frac{\gamma}2\tan\frac{ \alpha}2=1$.
I will use $\displaystyle a,b,c$ instead because it is easier to type in LaTeX.

If $\displaystyle a,b,c$ are angles of a triangle then $\displaystyle a+b+c = \pi$.

That means, $\displaystyle \tan \frac c2 = \tan \left( \frac{\pi - a - b}2 \right) = \tan \left( \frac{\pi}2 - \frac{a+b}2 \right) = \cot \frac{a+b}2$

Thus, $\displaystyle \sum_{cyc} \tan \frac a2 \tan \frac b2 = \tan \frac a2 \tan \frac b2 + \tan \frac a2 \cot \frac{a+b}2 + \tan \frac b2 \cot \frac{a+b}2$

Factor, $\displaystyle \tan \frac a2 \tan \frac b2 + \cot \frac{a+b}2 \left( \tan \frac a2 + \tan \frac b2 \right)$

Use identity, $\displaystyle \tan \frac a2 \tan \frac b2 + \left( \frac{1 - \tan \frac a2 \tan \frac b2}{\tan \frac a2 + \tan \frac b2} \right) \cdot \left( \tan \frac a2 + \tan \frac b2 \right) = \tan \frac a2 \tan \frac b2 + 1 - \tan \frac a2 \tan \frac b2 = 1$

3. Thank you very much, just a little typo at the end...
$\displaystyle \dots= \tan \frac a2 \tan \frac b2 + 1 - \tan \frac a2 \tan \frac b2 = 1$

4. Originally Posted by james_bond
Thank you very much, just a little typo at the end...
$\displaystyle \dots= \tan \frac a2 \tan \frac b2 + 1 - \tan \frac a2 \tan \frac b2 = 1$
Fixed.

5. Sorry but in a triange $\displaystyle a+b+c=\pi$. Am I wrong? .
(Of course the rest seems to be correct...)

6. Originally Posted by james_bond
Sorry but in a triange $\displaystyle a+b+c=\pi$. Am I wrong? .
(Of course the rest seems to be correct...)
Yes, another mistake. I meant they add up to $\displaystyle \pi$.