If the cos(x) = -12/13, and tan(x) < 0, find

sin(x)
tan(x)
sec(x

{Sin[x], Tan[x], Sec[x]} =

{5/13, -(5/12), -(13/12)}

tan(x) = sin(x)/cos(x)
tan(x) < 0 and cos(x) < 0 ⇒ sin(x) > 0
sin²(x) = 1 - cos²(x)
Since sin(x) > 0,
sin(x) = √(1 - cos²(x))
. . . . .= √(1 - 144/169)
. . . . .= √(25/169)
. . . . .= 5/13
tan(x) = sin(x)/cos(x)
. . . . .= (5/13)/(-12/13)
. . . . .= -5/12

sec(x)= 1/cos(x)
. . . . .= -13/12

make a right triangle of side -12, 13 and third side:

13^2-(-12)^2 = 5^2 third side is 5

thus sin x = 5/13

tan x = -12/ 5

sec x = -12/13

: )

5-12-13 triangle

sin x = 5/13

tan x = -5/12

sec x = -13/12

x = 157.38 degrees

sinx = 0.3846
tanx = - 0.416666666666
secx = - 1.0833333...........1/cosx