If tanθ + secθ = k, where k > 0, then cosθ written in terms of k is:

Because se$c^2$θ − ta$n^2$θ = 1, the conjugate of the given sum is secθ − tanθ $=1/k$. Adding the two equations gives 2secθ = $k+1/k=(k^2+1)/k$, so secθ = $(k^2+1)/(2k)$ and cosθ is its reciprocal, $2k/(k^2+1)$. Check with $k=3$: cosθ = 0.6, so θ ≈ 53.13° and tanθ + secθ = 1.333 + 1.667 = 3. Stopping one step early reports secθ = $(k^2+1)/(2k)$; subtracting the equations instead gives 2tanθ, and the resulting sinθ = $(k^2-1)/(k^2+1)$.