46) $$ \frac{\operatorname{sen} \alpha-\operatorname{sen} \beta}{\cos \alpha-\cos \beta}+\frac{\cos \alpha+\cos \beta}{\operatorname{sen} \alpha+\operatorname{sen} \beta}=0 $$

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1. Empezamos usando las fórmulas de transformación producto-suma:

$$
   \operatorname{sen} \alpha - \operatorname{sen} \beta = 2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)
   $$
   
   $$
   \cos \alpha - \cos \beta = -2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)
   $$
   
   $$
   \cos \alpha + \cos \beta = 2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)
   $$
   
   $$
   \operatorname{sen} \alpha + \operatorname{sen} \beta = 2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)
   $$

2. Sustituimos en la primera fracción:

$$
   \frac{\operatorname{sen} \alpha-\operatorname{sen} \beta}{\cos \alpha-\cos \beta} = \frac{2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)}{-2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)} = -\frac{\cos\left(\frac{\alpha+\beta}{2}\right)}{\sin\left(\frac{\alpha+\beta}{2}\right)} = -\cot\left(\frac{\alpha+\beta}{2}\right)
   $$

3. Sustituimos en la segunda fracción:

$$
   \frac{\cos \alpha+\cos \beta}{\operatorname{sen} \alpha+\operatorname{sen} \beta} = \frac{2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)}{2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)} = \frac{\cos\left(\frac{\alpha+\beta}{2}\right)}{\sin\left(\frac{\alpha+\beta}{2}\right)} = \cot\left(\frac{\alpha+\beta}{2}\right)
   $$

4. Sumamos ambas fracciones:

$$
   -\cot\left(\frac{\alpha+\beta}{2}\right) + \cot\left(\frac{\alpha+\beta}{2}\right) = 0
   $$

La expresión se simplifica a $$ 0 $$, lo que demuestra la identidad.
La expresión se simplifica a $$ 0 $$, lo que demuestra la identidad.

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