Could someone explain why this is the case? Or at least tell me which formula has been used. I do not understand how it is possible to take out one of the terms in the fraction contained in the argument of the arctan function. Thanks
Expressing the arctan function in a different form?
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trigonometry
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0Draw a picture. Make a right triangle with one leg $1$ and one leg $\omega$. – 2017-01-04
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$\omega = tan \theta$ for some $\theta$. Then $1/\omega = \cot \theta.$ COtangent is called COtangent because it's the tangent of the COmplementary angle. So $\arctan 1/\omega$ and $\arctan \omega$ are complementary angles. So they add to $\pi/2$.
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Here's a simple geometric approach for $\omega > 0$.
Consider a right triangle with non-right angles $\alpha$ and $\beta$. Without loss of generality, you can choose the adjacent side to angle $\alpha$ to be of length $1$; the length of the opposite side you call $\omega$. Clearly $\alpha+\beta = \tfrac{\pi}{2}$ and you have $\tan\alpha = \omega$ and $\tan \beta = \tfrac{1}{\omega}$, so $\alpha + \beta = \ldots$.