Step 1: Verify numerically first (sanity check):
$\arctan(0.2)\approx11.31^\circ$, $\arctan(1/7)\approx8.13^\circ$, $\arctan(1/3)\approx18.43^\circ$, $\arctan(1/8)\approx7.13^\circ$; these sum to $\approx45^\circ=\pi/4$, so the identity is plausible before proving it algebraically.
Step 2: Group differently — pair 1/5 with 1/3, and 1/7 with 1/8:
$\tan^{-1}\frac15+\tan^{-1}\frac13=\tan^{-1}\dfrac{1/5+1/3}{1-1/15}=\tan^{-1}\dfrac{8/15}{14/15}=\tan^{-1}\dfrac{8}{14}=\tan^{-1}\dfrac47$.
Step 3: Second pair:
$\tan^{-1}\frac17+\tan^{-1}\frac18=\tan^{-1}\dfrac{1/7+1/8}{1-1/56}=\tan^{-1}\dfrac{15/56}{55/56}=\tan^{-1}\dfrac{15}{55}=\tan^{-1}\dfrac3{11}$.
Step 4: Combine the two new results:
$\tan^{-1}\frac47+\tan^{-1}\frac3{11}=\tan^{-1}\dfrac{4/7+3/11}{1-12/77}=\tan^{-1}\dfrac{(44+21)/77}{65/77}=\tan^{-1}\dfrac{65}{65}=\tan^{-1}(1)=\dfrac\pi4$.
Final Answer:
Grouping the four terms differently still collapses to $\tan^{-1}(1)$, confirming the identity independent of pairing order.
\[ \boxed{\pi/4} \]