simplifying the trigonometric expression $\frac{\cot\theta - \tan\theta}{\sin\theta + \cos\theta} + \frac{1}{\cos\theta}$ I'm stuck on this expression for which I don't have any answer.
$$\frac{\cot\theta - \tan\theta}{\sin\theta + \cos\theta}+\frac{1}{\cos\theta}$$
Need some help on the simplifying steps.

Here is what I get from a calculator
 A: If you let $\cos\theta$ be $c$ and $\sin\theta$ be $s$ then you have:
$$ \frac{\frac{c}{s}-\frac{s}{c}}{s+c}+\frac{1}{c}=\frac{\frac{c^2}{s}-s+s+c}{(s+c)c}=\frac{\frac{cs+c^2}{s}}{sc+c^2} =\frac{cs+c^2}{s(cs+c^2)}=\frac{1}{s}$$
or it can be simplifed to:
$$\frac{1}{\sin\theta}$$
A: HINT
cancell s+c in num and den
$$\frac{c/s- s/c}{s+c}+\frac{1}{c}= ( c-s)/s c + ... $$
A: Here is a detailed answer:
$$
\require{cancel}
\begin{align}
\frac{\cot\theta-\tan\theta}{\sin\theta+\cos\theta}+\frac{1}{\cos\theta}&=\frac{\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}}{\sin \theta+\cos \theta}+\frac{1}{\cos \theta}\tag{1}\label{ko-eq1}\\
&=\frac{\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}}{\sin \theta+\cos \theta}+\frac{1}{\cos \theta}\tag{2}\label{ko-eq2}\\
&=\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}\cdot\frac{1}{\sin \theta+\cos \theta}+\frac{1}{\cos \theta}\tag{3}\label{ko-eq3}\\
&=\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}+\frac{1}{\cos \theta}\tag{4}\label{ko-eq4}\\
&=\frac{\cos^{2}\theta-\sin^{2}\theta+\sin\theta\left(\sin \theta+\cos \theta\right)}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}\tag{5}\label{ko-eq5}\\
&=\frac{\cos^{2}\theta\cancel{-\sin^{2}\theta}\cancel{+\sin^{2}\theta}+\sin\theta\cos \theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}\tag{6}\label{ko-eq6}\\
&=\frac{\cos^{2}\theta+\sin\theta\cos \theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}\tag{7}\label{ko-eq7}\\
&=\frac{\cancel{\cos\theta}\bcancel{\left(\sin \theta+\cos \theta\right)}}{\sin\theta\cancel{\cos\theta}\bcancel{\left(\sin \theta+\cos \theta\right)}}\tag{8}\label{ko-eq8}\\
&=\frac{1}{\sin\theta}\\
&=\csc\theta
\end{align}\\
$$
\eqref{ko-eq1}: Expand $\cot\theta$ and $\tan\theta$ as $\frac{\cos\theta}{\sin\theta}$ and $\frac{\sin\theta}{\cos\theta}$, respectively.
\eqref{ko-eq2}: Find common denominator of $\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}$.
\eqref{ko-eq3}: Rewrite $\frac{\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}}{\sin \theta+\cos \theta}$ as $\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}\cdot\frac{1}{\sin \theta+\cos \theta}$ and multiply.
\eqref{ko-eq4}: Multiplication resut.
\eqref{ko-eq5}: Find common denominator of $\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}+\frac{1}{\cos \theta}$ and continue.
\eqref{ko-eq6}: Cancel the terms.
\eqref{ko-eq7}: After cancellation. 
\eqref{ko-eq8}: Final cancellation and finish.
I hope this helps.
