If $x=a$·secθ with $a>0$ and θ acute, then √$(x^2-a^2)/x$ equals:

$x^2-a^2=a^2$(se$c^2$θ − 1) = $a^2$ta$n^2$θ, so √$(x^2-a^2)=a$·tanθ. Dividing by $x=a$·secθ gives tanθ/secθ = (sinθ/cosθ)·cosθ = sinθ. Cancelling the wrong way round produces the trap tanθ.