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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.

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

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    $\begingroup$ What's your different answer? Show your work? $\endgroup$ – Vincent May 23 '14 at 4:45
  • $\begingroup$ hint: once you get a reduced answer you can reduce it even further by $\sin 2\theta = 2\sin\theta\cos\theta$ $\endgroup$ – John Joy Aug 6 '14 at 14:56
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Remember that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ and $\sec(\theta) = \frac{1}{\cos(\theta)}$.

Substitute these into your original expression, cancel terms where appropriate, and simplify. It is also entirely possible that your answer and your teacher's answer are equivalent, which is something you'll have to look out for more and more as you progress in math. :)

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Remember that it's always possible your teacher is wrong. People make mistakes, and sometimes people are particularly prone to mistakes when they are trying to work in front of a group of students.

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 $$ \frac{\tan \theta \cos \theta}{\sec \theta} = \frac{(s/c)(c)}{(1/c)} = sc = \sin \theta \cos \theta. $$

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$$ \frac{\tan \theta \cos \theta}{\sec \theta} = \frac{\frac{\sin\theta}{\cos\theta}\cdot \cos\theta}{\frac{1}{\cos\theta}} = \cos\theta\left( \frac{\sin\theta}{\cos\theta}\cdot \cos\theta \right) = \sin\theta\cos\theta $$

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