Reducing Function to be Expressed in Terms of 1 Trigonometric Function 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 $x(y)$.

$\tan x + \sec x = \dfrac{\sin x + 1}{\cos x} = \dfrac{(\sin(\frac{x}{2}) + \cos(\frac{x}{2}))^2}{(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))(\cos(\frac{x}{2}) + \sin(\frac{x}{2}))} = \dfrac{\sin(\frac{x}{2}) + \cos(\frac{x}{2})}{\cos(\frac{x}{2}) - \sin(\frac{x}{2})} = \dfrac{\tan(\frac{x}{2}) + 1}{1 - \tan(\frac{x}{2})}$.
Let $t = \tan(\frac{x}{2})$, then:
$f(x) = f(t) = \dfrac{t + 1}{1 - t}$
