<s>[INST] 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.
 [/INST] 
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*}
</s>