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Is is possible to simplify the expression $\arctan(y)-\arctan(x)=c$.

I tried writing the expression in the form $\frac{\arcsin(y)}{\arccos(y)}-\frac{\arcsin(x)}{\arccos(x)}=c$ but it does not lead to anyting. How do I reduce it I want to eliminate $\arctan$ from the expression.

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    $\begingroup$ It wouldn't lead to anything, because what you wrote is plain wrong. $\endgroup$ – Kaster Feb 1 '16 at 4:53
  • $\begingroup$ Arctan $\ne $ arcsin/arccos. $\endgroup$ – fleablood Feb 1 '16 at 5:47
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Let $\theta_1 = \arctan (y)$ and $\theta_2 = \arctan (x)$, we have $$ \theta_1 - \theta_2 = c.$$ So $$ \tan(c) = \tan(\theta_1 - \theta_2) = \frac{\tan(\theta_1) - \tan(\theta_2)}{1 + \tan(\theta_1)\tan(\theta_2)} = \frac{y - x}{1 + xy}.$$

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