While solving a projectile motion problem, i found the elevation angle as

θ = arccos ((2/3) + (v^2/3gR)) (which is correct according the solution). The next step requires tan θ, which I calculated as equal to √(5 - (v^4/g^2.R^2) - (4v^2/gR))/ (2 + (v^2/gR)). The solution in the text says that tan θ = - √(1 - (v^2/gR))/(√(2 + (v^2/gR))). Can anyone help me on what I am missing?