Answer: A
Given \( \cot\alpha = \tan(\beta + \gamma) \) Then, \( \cot\alpha = \cot[90^\circ - (\beta + \gamma)] \) So, \( \alpha = 90^\circ - (\beta + \gamma) \) Therefore, \( \alpha + \beta + \gamma = 90^\circ \) \(\therefore\) \( \sin(\alpha + \beta + \gamma) = \sin 90^\circ = 1 \)
Given \( \cot\alpha = \tan(\beta + \gamma) \) Then, \( \cot\alpha = \cot[90^\circ - (\beta + \gamma)] \) So, \( \alpha = 90^\circ - (\beta + \gamma) \) Therefore, \( \alpha + \beta + \gamma = 90^\circ \) \(\therefore\) \( \sin(\alpha + \beta + \gamma) = \sin 90^\circ = 1 \)