Let

\arctan(1/3)=m<\dfrac{\pi}{4} ,

\arctan(1/7)=n<\dfrac{\pi}{4}

We get,

\tan m =\dfrac{1}{3}, \tan n=\dfrac{1}{7}

\tan 2m=\dfrac{2\tan m}{1-\tan^2 m}=\dfrac{2\cdotp \dfrac{1}{3} }{1-(\dfrac{1}{3})^2 } =\dfrac{3}{4}

\tan(2m+n)=\dfrac{\tan 2m+\tan n}{1-\tan 2m\cdotp \tan n } =\dfrac{\dfrac{3}{4}+\dfrac{1}{7} }{1-\dfrac{3}{4}\cdotp \dfrac{1}{7} } =1

Therefore, 2m+n=\dfrac{\pi}{4}

2\arctan \dfrac{1}{3} +\arctan\dfrac{1}{7} =\dfrac{\pi}{4}