Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 552

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\[1)\ 1 + tg\ a \bullet tg\ \beta = \frac{\cos(a - \beta)}{\cos a \bullet \cos\beta}\]

\[1 + tg\ a \bullet tg\ \beta =\]

\[= 1 + \frac{\sin a}{\cos a} \bullet \frac{\sin\beta}{\cos\beta} =\]

\[= \frac{\cos a \bullet \cos\beta + \sin a \bullet \sin\beta}{\cos a \bullet \cos\beta} =\]

\[= \frac{\cos(a - \beta)}{\cos a \bullet \cos\beta}\]

\[\frac{\cos(a - \beta)}{\cos a \bullet \cos\beta} = \frac{\cos(a - \beta)}{\cos a \bullet \cos\beta}\]

\[Что\ и\ требовалось\ доказать.\]

\[2)\ tg\ a - tg\ \beta = \frac{\sin(a - \beta)}{\cos a \bullet \cos\beta}\]

\[tg\ a - tg\ \beta = \frac{\sin a}{\cos a} - \frac{\sin\beta}{\cos\beta} =\]

\[= \frac{\sin a \bullet \cos\beta - \sin\beta \bullet \cos a}{\cos a \bullet \cos\beta} =\]

\[= \frac{\sin(a - \beta)}{\cos a \bullet \cos\beta}\]

\[\frac{\sin(a - \beta)}{\cos a \bullet \cos\beta} = \frac{\sin(a - \beta)}{\cos a \bullet \cos\beta}\]

\[Что\ и\ требовалось\ доказать.\]