Finding various things about $F(x)=\sec(x) + \tan(x)$ F:$[0,2\pi) - \{\frac{\pi}2, \frac{3\pi}2\}\rightarrow\mathbb R -\{0\})$
We need to find whether the function is bijective or not .

So $F'(x) = \sec x ( \sec x + \tan x )$ , which is obviously non-negative
 ($\frac{\pi}2 > x \ge 0$). So the function goes from 
 $1$ (at $x=0$) and $\rightarrow\infty$ (both $\sec x$ and $\tan x$ approaches infinity) as $x \rightarrow\frac{\pi}2$.
For $x$ in $(\frac{\pi}{2},π]$ again $F'(x)>0$ as all terms are negative (negative $\times$ negative $>0$), so the function goes from approaching $-\infty$ (at $\frac{π}{2}$) to $-1$ (at $\pi$).
From $x$ in 3rd quadrant $F'(x) = \frac{\sec x}{ \sec x - \tan x}$ which is positive as sec(x) is negative and tan(x) is positive . In fourth quadrant also as $\sec x$ is positive and $\tan x$ is negative $F'(x) >0$. 
$F(x)$ approaches zero as $x \rightarrow\frac{3\pi}2$ as $\sec x  + \tan x = \frac1{\sec x - \tan x }$ when $\sec x$ and $\tan x$ approaches $\infty$ or $-\infty$ but with opposite signs. So $F(x)$ goes from $-1$ (at $\pi$) to $+1$ (at $2\pi$).
Hence the function is bijective. What i want to know if there is a simpler to solve this question like $a \sin x  + b\cos x$ where you convert them to a simpler trigonometric expression. Any other methods are also welcome.
 A: The function is indeed bijective, for it only misses the value $0$.
If you write the function as
$$
F(x)=\frac{1+\sin x}{\cos x}
$$
then you have it is everywhere differentiable (hence continuous) with
$$
F'(x)=\frac{\cos^2x+(1+\sin x)\sin x}{\cos^2x}=\frac{1+\sin x}{\cos^2x}
$$
which is everywhere positive in the specified domain. Hence the function is strictly increasing everywhere.
Since $F(0)=1$ and
$$
\lim_{x\to(\pi/2)^-}F(x)=\infty,\quad
\lim_{x\to(\pi/2)^+}F(x)=-\infty,\quad
\lim_{x\to(3\pi/2)}F(x)=0,\quad
\lim_{x\to2\pi}F(x)=1
$$
we get that the only missed value is $0$.
A: Using the definitions of sec and tan we have:
$$
\begin{align}
f(x) &= \sec x + \tan x\\
&= \frac1{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x}
\end{align}
$$
We can fix the singularity at $x_1=3\pi/2$ as follow, for example by applying de l'Hôpital in (1):
$$f(x_1) = \lim_{x\to 3\pi/2} f(x) \stackrel{(1)}= \frac{\cos (3\pi/2)}{-\sin (3\pi/2)} = 0$$
Using the quotient rule to get the derivative of ƒ:
$$
\begin{align}
f'(x) &= \frac{\cos^2 x + (1+\sin x)\sin x}{\cos^2 x}\\
&\stackrel{(2)}=  \frac{1+\sin x}{\cos^2 x} 
 \stackrel{(2)}= \frac{1 + \sin x}{1-\sin^2 x}\\
&\stackrel{(3)}= \frac1{1-\sin x}
\end{align}
$$
Where (2) uses $\sin^2+\cos^2=1$ and (3) is shortening out $1+\sin x$ which is non-zero for $x\neq x_1$.  The derivative has a pole at $x_0=\pi/2$ and is strictly positive everywhere else, hence ƒ is injective in [0,π/2) and injective in (π/2,2π).
Because $f(0) = 1$ and $f(x_0^-)=\infty$ we have $f([0,\tfrac\pi 2))=[1,\infty)$.
Because $f(x_0^+)=-\infty$ and $f(2\pi^-)=f(0)=1$ we have $f((\tfrac\pi 2,1)) = (-\infty,1)$.
This means ƒ maps $[0,2\pi)\!\setminus\!\{\tfrac\pi 2\}$ uniquely onto $\mathbb R$ provided we fixed the singularity at 3π/2.
If we remove 3π/2 then we have to remove 0 from the image of ƒ and we get:
$$ f \text{ maps } [0,2\pi)\!\setminus\!\{\tfrac{\pi}2,\tfrac{3\pi}2\} \text{ uniquely onto } \mathbb R\!\setminus\! 0 $$
Note: We can also fix the singularity of the derivative in 3π/2 as obvious, but this does not matter here. Things would be the same if the derivative had a pole at 3π/2.
