Trig Identity Proof of Tan(x) + Tan(y)
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.
$\endgroup$ 82 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$For fun, here's a picture-proof:
$$\begin{align} 2\;|\triangle OAB| = \qquad\qquad |\overline{OR}|\;|\overline{AB}| \;&=\; |\overline{OA}|\;|\overline{OB}|\;\sin\angle AOB \\[6pt] 1\cdot (\;\tan\alpha + \tan\beta\;) \;&=\; \sec\alpha \;\cdot\;\sec\beta\;\cdot \;\sin(\alpha+\beta) \\ \tan\alpha + \tan\beta \;&=\; \frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta} \end{align}$$
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