Ii
MATU • Trigonometria
MATU_TRI_497
Libro de Trigonometría
Enunciado
Si $A + B + C = \frac{\pi}{2}$, demuestre que:
$$ \cot A + \cot B + \cot C = \cot A \cot B \cot C $$
$$ \cot A + \cot B + \cot C = \cot A \cot B \cot C $$
Solución Paso a Paso
1. Desarrollo:
Partimos de $A + B = \frac{\pi}{2} - C$.
Aplicamos cotangente:
$$ \cot(A + B) = \cot\left(\frac{\pi}{2} - C\right) $$
Usamos la identidad $\cot(\frac{\pi}{2} - \theta) = \tan \theta$:
$$ \frac{\cot A \cot B - 1}{\cot A + \cot B} = \tan C = \frac{1}{\cot C} $$
Multiplicando en cruz:
$$ \cot C(\cot A \cot B - 1) = \cot A + \cot B $$
$$ \cot A \cot B \cot C - \cot C = \cot A + \cot B $$
$$ \cot A \cot B \cot C = \cot A + \cot B + \cot C $$
2. Conclusión:
$$ \boxed{\cot A + \cot B + \cot C = \cot A \cot B \cot C} $$
Partimos de $A + B = \frac{\pi}{2} - C$.
Aplicamos cotangente:
$$ \cot(A + B) = \cot\left(\frac{\pi}{2} - C\right) $$
Usamos la identidad $\cot(\frac{\pi}{2} - \theta) = \tan \theta$:
$$ \frac{\cot A \cot B - 1}{\cot A + \cot B} = \tan C = \frac{1}{\cot C} $$
Multiplicando en cruz:
$$ \cot C(\cot A \cot B - 1) = \cot A + \cot B $$
$$ \cot A \cot B \cot C - \cot C = \cot A + \cot B $$
$$ \cot A \cot B \cot C = \cot A + \cot B + \cot C $$
2. Conclusión:
$$ \boxed{\cot A + \cot B + \cot C = \cot A \cot B \cot C} $$