Reducing Function to be Expressed in Terms of 1 Trigonometric Function

Question

I am looking for a way to express $\tan(x) + \sec(x)$ as a function expressed in terms of a single trigonometric function. So far I have it down to:

$$ \frac{\sqrt{1 - \cos^2 x} + 1}{\cos(x)} $$

Is there a more clean way to define this. Basically I have:

$$ y = \frac{\sqrt{1 - \cos^2 x} + 1}{\cos(x)} $$

and I need to express $x(y)$.

Answer 1

$$ \begin{align} & \tan x + \sec x = \frac{\sin x + 1}{\cos x} = \frac{\cos y + 1}{\sin y} = \frac{1}{\left( \frac{\sin y}{1+\cos y} \right)} = \frac{1}{\tan\frac y 2} \\[15pt] = {} & \cot\frac y 2 = \cot\left(\frac{\frac\pi2 - x}{2}\right) = \cot\left(\frac\pi4-\frac x2\right) \end{align} $$ where $y=\dfrac\pi2-x$.

One of many tangent half-angle formulas says $$ \tan x+\sec x = \tan\left(\frac\pi4 + \frac x 2\right). $$ Privately I think of this as the "cartographer's tangent half-angle formula" because of the occurrence of the function $x\mapsto\log(\tan x+\sec x)$ in the Mercator projection. The Mercator projection is characterized ("characterized" in the precisely defined mathematical sense of the word) by the fact that that compass bearings correspond to directions on the map, so that, for example $13^\circ$ east of north on the earth always corresponds to $13^\circ$ clockwise from straight up on the map, regardless of geographic location.

Answer 2

Hint: $$\sin^2(x)+\cos^2(x) = 1 \Rightarrow 1-\cos^2(x) = \sin^2(x)$$

Answer 3

$\tan x + \sec x = \dfrac{\sin x + 1}{\cos x} = \dfrac{(\sin(\frac{x}{2}) + \cos(\frac{x}{2}))^2}{(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))(\cos(\frac{x}{2}) + \sin(\frac{x}{2}))} = \dfrac{\sin(\frac{x}{2}) + \cos(\frac{x}{2})}{\cos(\frac{x}{2}) - \sin(\frac{x}{2})} = \dfrac{\tan(\frac{x}{2}) + 1}{1 - \tan(\frac{x}{2})}$.

Let $t = \tan(\frac{x}{2})$, then:

$f(x) = f(t) = \dfrac{t + 1}{1 - t}$