# Simplifying trigonometric expression

#### otownsend

Hi,

I hope someone can help me properly simplify the following expression:

The solution is apparently

Sincerely,
Olivia

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#### SlipEternal

MHF Helper
\begin{align*}\dfrac{\sin(f+g)+\sin(f-g)}{\cos(f+g) + \cos(f-g)} & = \dfrac{\sin f \cos g + \cancel{\sin g \cos f} + \sin f \cos g + \cancel{- \sin g \cos f}}{\cos f \cos g + \cancel{-\sin f \sin g} + \cos f \cos g + \cancel{\sin f \sin g}} \\ & = \dfrac{\cancel{2}\sin f \cancel{\cos g}}{\cancel{2}\cos f \cancel{\cos g}} \\ & = \tan f\end{align*}

Because we cancelled $\cos g$ from top and bottom, we need $\cos g \neq 0$. Because $\tan f$ is not defined when $\cos f = 0$, there is no need to specifically identify that as a condition.

You just missed the last steps where you cancel the 2's and cancel the $\cos g$ from top and bottom.

#### otownsend

Ok that makes perfect sense. Thanks for you help!