Trig Equation with Reciprocals?

When θ is in the fourth quadrant, i.e. 270° < θ < 360° :
sinθ < 0

secθ = √5
1/cosθ = √5
cosθ = 1/√5

sin²θ + cos²θ = 1
sin²θ + (1/√5)² = 1
sin²θ + (1/5) = 1
sin²θ = 4/5

As sinθ < 0, then sinθ = -√(4/5)
sinθ = -√(20/25)
sinθ = -2(√5)/5

secθ = √5
1/cosθ = √5
cosθ = 1/√5
sin²θ = 1 - cos²θ = ⅘
sinθ = -√(⅘) = -2/√5

OK, since sec is positive in quadrant IV, well take the positive value. How does that sound?

secθ = √5
1/cosθ = √5
cosθ = 1/√5

Now use the Pythagorean identity :
sin^2 (θ) + cos^2(θ) = 1

sinθ = √(1 - cos^2 (θ))
sinθ = √(1 - 1/5)
sinθ = √(4/5)
sinθ = 2/√5

Hope this helps :)