Найдите :
sinα, tgα, ctgα,
если

cosα= - 5/13
π/2 <α< π

1)

sin²α + cos²α = 1

sin²α = 1 - cos²α = 1 - (-5/13)² = 1 - 25/169 = 144/169

Так как π/2 < α < π, то sin α > 0, тогда

sin α = √(144/169) = 12/13.

2)

tg α = sin α/cos α = 12/13 : (-5/13) = - (12•13)/(13•5) = - 12/5 = - 2 2/5.

3)

ctg α = 1/tg α = 1 : (-12/5) = - 5/12.