Question

3 задание найдите tg a

Answer 1

$sina=-12/13,\; \; a\in IV\\\\cosa=\pm\sqrt{1-sin^2a}=\pm\sqrt{1-(-12/13)^2}=\\=\pm\sqrt{1-144/169}=\pm\sqrt{25/169} =\pm5/13\\a\in IV\; \; => cosa=5/13\\\\tga=sina:cosa\\tga=-12/13:5/13 = -12/5 =-2,4$