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

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    $\begingroup$ I'm not sure if I understand you correctly, do you mean $\frac{sec \theta}{(sec\theta)(tan\theta)} $ on the LHS of the first row? $\endgroup$ – BigbearZzz Oct 4 '15 at 12:13
  • $\begingroup$ Please check that the statement of the identity is correct. Also, please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 4 '15 at 12:14
  • $\begingroup$ @BigbearZzz yes $\endgroup$ – iAthena Oct 4 '15 at 12:17
  • $\begingroup$ Try some special angle $\theta$ that you are familiar with. $\endgroup$ – André Nicolas Oct 4 '15 at 12:38

It's wrong.

Here's why:


\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.


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.


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