Найти tg a, если sin a = -4√41/41 , [π; 3π/2]

[latex]sin a = -\frac{4\sqrt{41}}{41}\\cosa=б\sqrt{1-sin^2a}=б\sqrt{1- (-\frac{4\sqrt{41}}{41})^2}=б\sqrt{1- \frac{16}{41}}=б\frac{5}{\sqrt{41}}=\\=б\frac{5\sqrt{41}}{41}\\[/latex] т.к. a в 3 четв. и cos в 3 чет отриц то [latex]cosa=-\frac{5\sqrt{41}}{41}\\tga=\frac{ -\frac{4\sqrt{41}}{41}}{-\frac{5\sqrt{41}}{41}}=\frac{4}{5}[/latex]