Establish the identity $\frac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta\$ Establish the identity:
$$\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta$$
The first step I got was:
$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\theta\big)}{\cos\theta + \tan\theta}$$
Then it tells me to rewrite the factor $$\cos\theta + \tan\theta$$
in the numerator using reciprocal identities.
How would I do that?
Here is what the assignment looked like:



 A: Hint: 
$$\cot\theta + \sec\theta = \frac{1}{\tan\theta} + \frac{1}{\cos\theta} = \frac{\cos\theta + \tan\theta}{\tan\theta\cos\theta}$$
A: On the one hand
\begin{align}
\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} &= 
\dfrac{\dfrac{\cos\theta}{\sin\theta} + \dfrac{1}{\cos\theta}}{\cos\theta + \dfrac{\sin\theta}{\cos\theta}} =
\dfrac{\cos^2\theta+\sin\theta}{\sin\theta\cos\theta} \div\dfrac{\cos^2\theta+\sin\theta}{\cos\theta}
\\ &=
\dfrac{\cos^2\theta+\sin\theta}{\sin\theta\cos\theta} \cdot \dfrac{\cos\theta}{\cos^2\theta+\sin\theta}\\ &= \dfrac{1}{\sin\theta}.
\end{align}
On the other, 
\begin{align}
\sec\theta\cot\theta &= \dfrac{1}{\cos\theta}\cdot\dfrac{\cos\theta}{\sin\theta} = \dfrac{1}{\sin\theta} .
\end{align}
Thus we have shown that 
$$\bbox[1ex, border:2pt solid #e10000]{\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta}$$
A: Continuing from what you got:
$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\theta\big)}{\cos\theta + \tan\theta}$$
and since $\sec \theta \cos \theta = 1, \cot \theta \tan \theta = 1$, expand the brackets:
$$\sec\theta \cot\theta = \frac{\cot \theta + \sec \theta}{\cos \theta + \tan \theta}$$
There is no need to rewrite $\sec \theta \cot \theta$.
