Question

помощь надо срочно адалаовлв

Answer 1

Ответ: tgα=-√7/3.

Пошаговое объяснение:

$\displaystyle\\cos\alpha =\frac{3}{4}\ \ \ \ \ \ \ \pi < \alpha < 2\pi \ \ \ \ \ \ tg\alpha =?\\\\\\sin^2\alpha +cos^2\alpha =1\\\\\\sin^2\alpha =1-cos^2\alpha =1-(\frac{3}{4} )^2=1-\frac{9}{16} =\frac{1*16-9}{16}=\frac{7}{16};\\\\\\sin\alpha =б\sqrt{\frac{7}{16} } =б\frac{\sqrt{7} }{4}.\\\\\\\pi < \alpha < 2\alpha \ \ \ \ \ \ \ \Rightarrow\\\\\\sin\alpha =-\frac{\sqrt{7} }{4} .\\\\\\tg\alpha =\frac{sin\alpha }{cos\alpha } =\frac{-\frac{\sqrt{7} }{4} }{\frac{3}{4} } =-\frac{\sqrt{7} }{3}.$