Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1036

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\[1)\ \frac{tg\ 29{^\circ} + tg\ 31{^\circ}}{1 - tg\ 29{^\circ} \bullet tg\ 31{^\circ}} =\]

\[= tg(29{^\circ} + 31{^\circ}) = tg\ 60{^\circ} = \sqrt{3}\]

\[2)\ \frac{\text{tg}\frac{7\pi}{16} - tg\frac{3\pi}{16}}{1 + tg\frac{7\pi}{16} \bullet tg\frac{3\pi}{16}} =\]

\[= tg\left( \frac{7\pi}{16} - \frac{3\pi}{16} \right) = tg\frac{4\pi}{16} =\]

\[\text{=}\text{tg}\frac{\pi}{4} = 1\]

\[3)\ \frac{1 + tg\ 10{^\circ} \bullet tg\ 55{^\circ}}{tg\ 55{^\circ} - tg\ 10{^\circ}} =\]

\[= \frac{1}{\text{tg}(55{^\circ} - 10{^\circ})} = \frac{1}{tg\ 45{^\circ}} =\]

\[= \frac{1}{1} = 1\]

\[4)\ \frac{1 - tg\ 13{^\circ} \bullet tg\ 17{^\circ}}{tg\ 17{^\circ} + tg\ 13{^\circ}} =\]

\[= \frac{1}{\text{tg}(17{^\circ} + 13{^\circ})} = \frac{1}{tg\ 30{^\circ}} =\]

\[= 1\ :\frac{1}{\sqrt{3}} = \sqrt{3}\]