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I am trying to prove the identity below to help with the simplification of another function that I'm investigating as it doesn't appear to be a standard trig identity.

$$ \tan\left(x\right) + \tan\left( y \right) = \frac{{\sin\left( {x + y} \right)}}{{\cos\left( x \right)\cos\left( y \right)}} $$

Any assistance gratefully appreciated.

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is your question a solution to formular 1, or a derivation of the identity in formular 2? – NicoDean Jul 19 '14 at 10:17
Essentially the first equation can be arranged to be either of the other two using the identity identified, but for the life of me I can't prove it and I wonder if that's because I'm either rusty or perhaps there are other simplifications coming into play that I'm missing... – ASBO Allstar Jul 19 '14 at 10:20
So you want a prove of that identity in formular 2? – NicoDean Jul 19 '14 at 10:26
Yes, I'm struggling to rearrange the first formula to be both the third and fourth formula, using the identity (formula 2). I'm unsure if I'm either missing simplifications along the way (perhaps other trig identities are needed) or my Maths needs a little more oil on the cogs. Thanks. – ASBO Allstar Jul 19 '14 at 10:32
I've stripped the question back even further and accepted your answer, thank you for your help.... Once my reputation is higher I'll come back and vote your answer up too... :) – ASBO Allstar Jul 19 '14 at 11:03
up vote 14 down vote accepted

You are asking about a proof of the identity $$ \tan\left(x\right) + \tan\left( y \right) = \frac{{\sin\left( {x + y} \right)}}{{\cos\left( x \right)\cos\left( y \right)}} $$

Using $\tan(x)=\frac{\sin(x)}{\cos(x)}$, we get $$\tan\left(x\right) + \tan\left( y \right) = \frac{\sin(x)}{\cos(x)} + \frac{\sin(y)}{\cos(y)}\\ =\frac{\sin(x)\cdot \cos(y) + \sin(y)\cdot \cos(x)}{\cos(x)\cdot \cos(y)}$$

Using the identity $\sin(x+y)=\sin(x)\cdot \cos(y) + \sin(y)\cdot \cos(x)$ gives you the answer of your question.

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For fun, here's a picture-proof:

enter image description here

$$\begin{align} 2\;|\triangle OAB| = \qquad\qquad |\overline{OR}|\;|\overline{AB}| \;&=\; |\overline{OA}|\;|\overline{OB}|\;\sin\angle AOB \\[6pt] 1\cdot (\;\tan\alpha + \tan\beta\;) \;&=\; \sec\alpha \;\cdot\;\sec\beta\;\cdot \;\sin(\alpha+\beta) \\ \tan\alpha + \tan\beta \;&=\; \frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta} \end{align}$$

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