Question

6. Найти tgа, если cosa = -2/3 и п 1) корень3/2
2) корень5/4
3) 1/2
4) 2корень из 3

Answer 1

$cosa=-\dfrac{2}{3}\\\\\star \ \ 1+tg^2a=\dfrac{1}{cos^2a}\ \ \Rightarrow \ \ \ tg^2a=\dfrac{1}{cos^2a}-1\ \ ,\ \ tga=\pm \sqrt{\dfrac{1}{cos^2a}-1}\ \ \star \\\\\\tga=\pm \sqrt{\dfrac{1}{4/9}-1}=\pm \sqrt{\dfrac{9}{4}-1}=\pm \sqrt{\dfrac{5}{4}}=\pm \dfrac{\sqrt5}{2}$

$1)\ \ a\in (\, 0\, ;\dfrac{\pi}{2})\ \ ili\ \ a\in (\, \pi \, ;\, \dfrac{3\pi}{2})\ \ \Rightarrow \ \ \ tga=+\dfrac{\sqrt5}{2}\\\\\\2)\ \ a\in (\, \dfrac{\pi}{2}\, ;\, \pi \, )\ \ ili\ \ a\in (\, \dfrac{3\pi}{2}\, ;\, 2\pi \, )\ \ \Rightarrow \ \ \ tga=-\dfrac{\sqrt5}{2}$

Answer 2

cosa=-2/3 → sina=√(1-cos²a) → sina=√(1-4/9) → sina=√5/3

tga=sina/cosa → tga=(√5/3):(-2/3) → tga=-√5/2