Express the function $f(x)= (1-\sin x)/(1+\sin x)$ as the sum of an even and odd function. Express the function $f(x)= (1-\sin x)/(1+\sin x)$ as the sum of an even and odd function.
 A: Hint: Given any $f(x),$ you have
$$f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2},$$
where the first is even and the second is odd. In applying this one has to make sure the domains are OK, so that has to be checked for your example.
A: $f(x)= \frac{1-\sin x}{1+\sin x} = \frac{(1-\sin x)^2}{(1+\sin x)\times (1-\sin x)} = \frac{1+\sin^2 x-2sinx}{\cos^2 x} = \frac{1}{\cos^2 x}+\tan^2 x  -2 \tan x \sec x  = sec^2 x+\tan^2 x  -2 \tan x \sec x$
Let the even function be $sec^2 x+\tan^2 x $
Let the odd function be $-2 \tan x \sec x $
Because the parity of $\sec^2 x$ and  $\tan^2 x$ and $\tan x$ and $\sec x $, it is easy to verify the parity of the above functions.
A: 
Hint: In order to answer such questions the systematic approach provided by @coffeemath is very helpful,  since it  provides you  a method to always find the odd and even part of a function $f$ regardless of its shape.

You want to split the function $f$ into the sum of an even function $g$ and and odd function $h$
\begin{align*}
f(x)=g(x)+h(x)
\end{align*}

Recall a function $g$ is even if $g(x)=g(-x)$.

If we put
\begin{align*}
g(x)=\frac{f(x)+f(-x)}{2}\tag{1}
\end{align*}
we obtain 
\begin{align*}
g(-x)=\frac{f(-x)+f(x)}{2}=g(x)
\end{align*} 

Recall a function $h$ is odd if $h(x)=-h(-x)$.

If we put
\begin{align*}
h(x)=\frac{f(x)-f(-x)}{2}\tag{2}
\end{align*}
we obtain
\begin{align*}
-h(-x)=-\frac{f(-x)-f(x)}{2}=\frac{f(x)-f(-x)}{2}=h(x)
\end{align*}

Putting all together results in
  \begin{align*}
g(x)+h(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}=f(x)
\end{align*}
  and we are done.

$$ $$

The even  part of $f(x)$
With the help of (1) we systematically find the even part
  \begin{align*}
g(x)&=\frac{1}{2}\left(f(x)+f(-x)\right)\\
&=\frac{1}{2}\left(\frac{1-\sin(x)}{1+\sin(x)}+\frac{1-\sin(-x)}{1+\sin(-x)}\right)\\
&=\frac{1}{2}\left(\frac{1-\sin(x)}{1+\sin(x)}+\frac{1+\sin(x)}{1-\sin(x)}\right)\\
&=\frac{1}{2}\cdot\frac{\left(1-\sin(x)\right)^2+\left(1+\sin(x)\right)^2}{1-\sin^2(x)}\\
&=\frac{1}{2}\cdot\frac{2+2\sin^2(x)}{1-\sin^2(x)}\\
&=\frac{1+\sin^2(x)}{\cos^2(x)}\\
\end{align*}
The odd part of $f(x)$
Similarly with the help of (2) we find the odd part
  \begin{align*}
h(x)&=\frac{1}{2}\left(f(x)-f(-x)\right)\\
&=\frac{1}{2}\left(\frac{1-\sin(x)}{1+\sin(x)}-\frac{1-\sin(-x)}{1+\sin(-x)}\right)\\
&=\frac{1}{2}\left(\frac{1-\sin(x)}{1+\sin(x)}-\frac{1+\sin(x)}{1-\sin(x)}\right)\\
&=\frac{1}{2}\cdot\frac{\left(1-\sin(x)\right)^2-\left(1+\sin(x)\right)^2}{1-\sin^2(x)}\\
&=\frac{1}{2}\cdot\frac{-4\sin(x)}{1-\sin^2(x)}\\
&=-\frac{2\sin(x)}{\cos^2(x)}\\
\end{align*}

We conclude the representation by even and odd part of the function $f$ is
\begin{align*}
f(x)=\frac{1+\sin^2(x)}{\cos^2(x)}-\frac{2\sin(x)}{\cos^2(x)}
\end{align*}
