Show that tan⁻¹1/3 + tan⁻¹1/5 + tan⁻¹1/7 + tan⁻¹1/8 is a simple angle…

Step 1. The first pair. $\dfrac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{15}} = \dfrac{8/15}{14/15} = \dfrac{4}{7}$. The sum is $\tan^{-1}\dfrac{4}{7} = 0.5191$.

Step 2. The last pair. $\dfrac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{56}} = \dfrac{15/56}{55/56} = \dfrac{3}{11}$. The sum is $\tan^{-1}\dfrac{3}{11} = 0.2663$.

Step 3. The two together. The product $\dfrac{4}{7} \times \dfrac{3}{11} = \dfrac{12}{77} < 1$. The fraction is $\dfrac{\frac{4}{7} + \frac{3}{11}}{1 - \frac{12}{77}} = \dfrac{65/77}{65/77} = 1$. The total is $\tan^{-1}1 = \dfrac{\pi}{4} = 0.7854$.