Sum of two arctangents

Text

Prove that $\arctan\dfrac12+\arctan\dfrac13=\dfrac{\pi}{4}$.

Solution

Set $\alpha=\arctan\frac12$ and $\beta=\arctan\frac13$, so $\tan\alpha=\frac12$, $\tan\beta=\frac13$. By the addition formula: $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}=\frac{\frac56}{\frac56}=1.$ Since $\alpha,\beta\in\left(0,\frac{\pi}{4}\right)$, the sum $\alpha+\beta$ lies in $\left(0,\frac{\pi}{2}\right)$; the only angle in that interval with tangent $1$ is $\frac{\pi}{4}$. Hence $\arctan\tfrac12+\arctan\tfrac13=\boxed{\dfrac{\pi}{4}}\approx 0.7854\ \text{rad}.$