Question

if 2sinx=1,π÷2<x<π and √2cosy=1,
3π÷2<y<2π, find the value of
tanx + tany÷ cosx - cosy

Answer 1

sinx = 1/2
cosx = -√3/2

siny = -1/√2
cosy = 1/√2

Now just plug in your values:

(-1/√3 - 1)/(-√3/2 - 1/√2) =

I'll let you massage that to a more pleasant appearance.

Answer 2

if 2sinx =1 ,π/2<x<π also √2 ×cosy=1, 3π/2<y<2π, find the value of tanx+tany÷cosx-cosy