If sinx = -3/5, tan x > 0, sec y = -13/5 and cot y < 0, find the following:

Question

a. csc(x + y)
b. sec (x - y)

How do I get the answers of
a. -65/33
b. -65/16

Thank you

Answer 1

from "sinx = -3/5, tan x > 0, sec y = -13/5 and cot y < 0 "
we can tell that x must be in quad III, and y must be in quad II

so if sinx - -3/5, then cosx = -4/5 ( I recognized the 3,4,5 right-angled triangle)
if secy = -13/5, then cosy = -5/13
and siny = 12/13 (I recognized the 5,12,13 triangle)

a) csc(x+y) = 1/sin(x+y)
so we'll find sin(x+y), then flip the answer

sin(x+y) = sinxcosy + cosxsiny
= (-3/5)(-5/13) + (-4/5)(12/13)
= 15/65 - 48/65
= -33/65

so csc(x+y) = -65/33

b) you try that one
remember sec(x+y) = 1/cos(x+y)
and cos(x+y) = cosxcosy - sinxsiny