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Решите задачу:

$1)\; \; \frac{tg \frac{\pi}{9}+tg \frac{5\pi}{36} }{1+tg \frac{31\pi}{36}\cdot tg \frac{\pi}{9} } = \frac{tg \frac{\pi}{9} +tg(\pi - \frac{31\pi}{36}) }{1+tg \frac{31\pi}{36} \cdot tg\frac{\pi}{9} } = \frac{tg \frac{\pi}{9} -tg \frac{31\pi}{36} }{1+tg \frac{31\pi}{36}\cdot tg \frac{\pi}{9} } =tg( \frac{\pi }{9} -\frac{31\pi}{36} )=\\\\=tg(-\frac{27\pi}{36})=-tg\frac{27\pi}{36}=-tg \frac{3\pi}{4}=-tg(\pi -\frac{\pi}{4})=tg \frac{\pi}{4}=1$

$2)\; \; tgx=-4\; ,\; \; -90^\circ \ \textless \ x\ \textless \ 90^\circ \\\\tgx\ \textless \ 0\; \; i\; \; -90^\circ\ \textless \ x\ \textless \ 90^\circ \; \; \; \Rightarrow \; \; \; x\in 4\; chetverti\; \; \to \; \; sinx\ \textless \ 0\\\\1+ctg^2x=\frac{1}{sin^2x}\; ,\; \; \to \; \; \; sin^2x=\frac{1}{1+ctg^2x}= \frac{1}{1+\frac{1}{tg^2x}} = \frac{tg^2x}{1+tg^2x} \\\\sin^2x= \frac{(-4)^2}{1+(-4)^2} = \frac{16}{1+16} = \frac{16}{17} \\\\Tak\; kak\; \; sinx\ \textless \ 0\; ,\; \; to\; \; sinx=-\sqrt{sin^2x}=-\sqrt{ \frac{16}{17}}=-\frac{4}{\sqrt{17}} =-\frac{4\sqrt{17}}{17}$