# How do you simplify $\frac{\tan \theta \cos \theta}{\sec \theta}$?

How do you solve/simplify this? I am having trouble solving for the correct answer. We did this in class and I am getting a different answer than what the teacher said it was.

$$\frac{\tan \theta \cos \theta}{\sec \theta}$$?

• hint: once you get a reduced answer you can reduce it even further by $\sin 2\theta = 2\sin\theta\cos\theta$ – John Joy Aug 6 '14 at 14:56
Remember that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ and $\sec(\theta) = \frac{1}{\cos(\theta)}$.
Use \begin{align*} \tan \theta &= \frac{\sin\theta}{\cos \theta} \\ \sec \theta &= \frac{1}{\cos \theta} \end{align*} So if $c = \cos \theta$, $s = \sin \theta$, we have $$\frac{\tan \theta \cos \theta}{\sec \theta} = \frac{(s/c)(c)}{(1/c)} = sc = \sin \theta \cos \theta.$$
$$\frac{\tan \theta \cos \theta}{\sec \theta} = \frac{\frac{\sin\theta}{\cos\theta}\cdot \cos\theta}{\frac{1}{\cos\theta}} = \cos\theta\left( \frac{\sin\theta}{\cos\theta}\cdot \cos\theta \right) = \sin\theta\cos\theta$$