Identity with arctan

Statement

Verify the notable identity $\arctan x + \arctan(1/x) = \dfrac{\pi}{2}$ for $x>0$. (Hint: let $\alpha = \arctan x$ and show that $\tan(\pi/2-\alpha) = 1/x$.)

Solution

Set $\alpha = \arctan x$; since $x>0$, we have $\alpha\in\left(0,\tfrac{\pi}{2}\right)$ and $\tan\alpha = x$. Using the complementary arc: $\tan\!\left(\tfrac{\pi}{2}-\alpha\right) = \frac{1}{\tan\alpha} = \frac{1}{x}.$ Since $\tfrac{\pi}{2}-\alpha\in\left(0,\tfrac{\pi}{2}\right)$, applying $\arctan$ (one-to-one on this interval): $\tfrac{\pi}{2}-\alpha = \arctan\!\left(\tfrac{1}{x}\right) \;\Rightarrow\; \arctan x + \arctan\!\left(\tfrac{1}{x}\right) = \boxed{\tfrac{\pi}{2}}. \quad\blacksquare$

Links

Topics: Goniometry
Concepts: Associated arcs · Arctangent
Functions: Arctangent · Inverse trigonometric functions
Skills: Proving · Using formulae
Exercise types: Proof