Question

Знайдіть tgα, якщо cosα = √ ̅1̅0̅ /10, α∊ (3π/2; 2π).

Answer 1

$\displaystyle\bf\\Cos\alpha =\frac{\sqrt{10} }{10} =\frac{\sqrt{10} }{(\sqrt{10} )^{2} } =\frac{1}{\sqrt{10} } \\\\\\\alpha \in\Big(\frac{3\pi }{2} \ ; \ 2\pi \Big)$

α - угол четвёртой четверти , значит  tgα < 0 .

$\displaystyle\bf\\1+tg^{2} \alpha =\frac{1}{Cos^{2} \alpha } \\\\\\tg^{2} \alpha =\frac{1}{Cos^{2}\alpha } -1=\frac{1}{(\frac{1}{\sqrt{10} })^{2} } -1=\frac{1}{\frac{1}{10} } -1=10-1=9\\\\\\tg\alpha =-\sqrt{9} =-3\\\\\\Otvet \ : \ tg\alpha =-3$