Fundamental Identities I was given a task to prove:
$$\frac{\sec\theta}{\sec\theta\tan\theta} = \sec\theta(\sec\theta + \tan\theta)$$
Then I replaced them with their Ratio and Reciprocal Identities
\begin{align*}
\sec\theta & = \frac{1}{\cos\theta}\\
\tan\theta & = \frac{\sin\theta}{\cos\theta}
\end{align*}
so I came up with this:
$$\frac{\frac{1}{\cos\theta}}{\frac{1}{\cos\theta} \cdot \frac{\sin\theta}{\cos\theta}} = \frac{1}{\cos\theta}\left(\frac{1}{\cos\theta} + \frac{\sin
\theta}{\cos\theta}\right)$$
then,
$$\frac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}} = \frac{1}{\cos\theta}\left(1+\frac{\sin\theta}{\cos\theta}\right)$$
and I had the reciprocal,
$$\frac{1}{\cos\theta} \cdot \frac{\cos\theta}{\sin\theta} = \frac{1}{\cos\theta}\left(1+\frac{\sin\theta}{\cos\theta}\right)$$
and I don't know what to do next, can someone explain to me how? T_T
 A: It's wrong.
Here's why:
$$\text{LHS}=\frac{\sec\theta}{\sec\theta\tan\theta}=\frac1{\tan\theta}=\frac{\text{adjacent}}{\text{opposite}}\phantom{....}$$
\begin{align}
\text{RHS}&=\sec\theta(\sec\theta+\tan\theta) \\[0.5ex]
&=\frac1{\cos\theta}\left(\frac 1{\cos\theta}+\frac{\sin\theta}{\cos\theta}\right) \\[0.6ex]
&=\frac1{\cos\theta}\left(\frac{1+\sin\theta}{\cos\theta}\right) \\[0.6ex]
&=\frac{1+\sin\theta}{\cos^2\theta}=\frac{\text{hypotenuse}+\text{opposite}}{\text{adjacent}}
\end{align}
Now, for $\frac{\text{adjacent}}{\text{opposite}}$ to be equal to $\frac{\text{hypotenuse}+\text{opposite}}{\text{adjacent}}$, $``\text{opposite"}$ must be equal to $``\text{adjacent"}$ and $``\text{hypotenuse"}$ must be $0$. Since this cannot be true, I suppose it is wrong.
A: Well the identity you mentioned is not correct:
Let $\theta = \frac{\pi}{4}$. Them your identity becomes $$\frac{\sec{\frac{\pi}{4}}}{(\sec{\frac{\pi}{4}})(\tan{\frac{\pi}{4}})}=1$$But the result you mentioned gives a different answer.
