Задача 19. Найдите \({\text{tg}}\,\,\left( {\alpha -\beta } \right)\), если \({\text{ctg}}\,\,\left( {\alpha + \dfrac{\pi }{2}} \right) = 3,\,\,\,\,{\text{tg}}\,\left( {\pi -\beta } \right) = -2\)
Решение
\({\rm{ctg}}\left( {\alpha + \dfrac{\pi }{2}} \right) = 3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-{\rm{tg}}\alpha = 3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\rm{tg}}\alpha = -3.\)
\({\rm{tg}}\left( {\pi -\beta } \right) = -2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-{\rm{tg}}\beta = -2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\rm{tg}}\beta = 2.\)
Воспользуемся формулой тангенса разности:
\({\rm{tg}}\left( {\alpha -\beta } \right) = \dfrac{{{\rm{tg}}\alpha -{\rm{tg}}\beta }}{{1 + {\rm{tg}}\alpha {\rm{tg}}\beta }} = \dfrac{{-3-2}}{{1-3 \cdot 2}} = 1.\)
Ответ: 1.