Show that 2 tan⁻¹1/3 + tan⁻¹1/7 is a simple angle…

Step 1. The double angle. $\tan\left(2\tan^{-1}\dfrac{1}{3}\right) = \dfrac{2/3}{1 - 1/9} = \dfrac{2/3}{8/9} = \dfrac{3}{4} = 0.75$. Since $\dfrac{1}{3}$ is between $-1$ and $1$, $2\tan^{-1}\dfrac{1}{3} = \tan^{-1}\dfrac{3}{4} = 0.6435$.

Step 2. Add the third term. The product $\dfrac{3}{4} \times \dfrac{1}{7} = \dfrac{3}{28} < 1$, so $\tan^{-1}\dfrac{3}{4} + \tan^{-1}\dfrac{1}{7} = \tan^{-1}\dfrac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{28}} = \tan^{-1}\dfrac{25/28}{25/28} = \tan^{-1}1$.

Step 3. The value. $\tan^{-1}1 = \dfrac{\pi}{4} = 0.7854$.