ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1298

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\[1)\ \frac{tg\ \alpha + tg\ \beta}{ctg\ \alpha + ctg\ \beta} =\]

\[= (tg\ \alpha + tg\ \beta)\ :\left( \frac{1}{\text{tg\ α}} + \frac{1}{\text{tg\ β}} \right) =\]

\[= (tg\ \alpha + tg\ \beta)\ :\frac{tg\ \beta + tg\ \alpha}{tg\ \alpha \bullet tg\ \beta} =\]

\[= (tg\ \alpha + tg\ \beta) \bullet \frac{tg\ \alpha \bullet tg\ \beta}{tg\ \beta + tg\ \alpha} =\]

\[= tg\ \alpha \bullet tg\ \beta;\]

\[2)\ \left( \sin a + \cos a \right)^{2} + \left( \sin a - \cos a \right)^{2} =\]

\[= 2\sin^{2}a + 2\cos^{2}a =\]

\[= 2 \bullet \left( \sin^{2}a + \cos^{2}a \right) =\]

\[= 2 \bullet 1 = 2;\]

\[3)\ \frac{\sin{\left( \frac{\pi}{4} + a \right) - \cos\left( \frac{\pi}{4} + a \right)}}{\sin{\left( \frac{\pi}{4} + a \right) + \cos\left( \frac{\pi}{4} + a \right)}} =\]

\[= \frac{\sqrt{2}\sin a}{\sqrt{2}\cos a} = tg\ a;\]

\[4)\ \frac{\sin a + 2\sin\left( \frac{\pi}{3} - a \right)}{2\cos\left( \frac{\pi}{6} - a \right) - \sqrt{3}\cos a} =\]

\[= \frac{\sin a + 2 \bullet \left( \frac{\sqrt{3}}{2}\cos a - \frac{1}{2}\sin a \right)}{2 \bullet \left( \frac{\sqrt{3}}{2}\cos a + \frac{1}{2}\sin a \right) - \sqrt{3}\cos a} =\]

\[= \frac{\sin a + \sqrt{3}\cos a - \sin a}{\sqrt{3}\cos a + \sin a - \sqrt{3}\cos a} =\]

\[= \frac{\sqrt{3} \bullet \cos a}{\sin a} = \sqrt{3}\text{\ ctg\ a.}\]