π < θ < 3pi/2, so θ ∈ 3 quadrant,
so cosθ<0, tg θ>0
we know that sin²θ+cos²θ=1
cosθ=-√(1-sin²θ)=-√1-\((\frac{ \sqrt{3} }{2})\)²=-√(1-\(\frac{3}{4})\)=- √0.25=- 0.5
tgθ=\(\frac{sin \alpha }{cos \alpha }\)= - \(\frac{ \sqrt{3} }{2}: (- 0.5)\)=√3
P.s. θ=\(\alpha\), i can't put θ in formula, sorry
