# 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

• 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? – BigbearZzz Oct 4 '15 at 12:13
• Please check that the statement of the identity is correct. Also, please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Oct 4 '15 at 12:14
• @BigbearZzz yes – iAthena Oct 4 '15 at 12:17
• Try some special angle $\theta$ that you are familiar with. – André Nicolas Oct 4 '15 at 12:38

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.

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.