Parag is working on a school project on right-angled triangles. He draws a right-angled triangle…

In right-angled triangle PQR with $\angle R = 90^\circ$, by Pythagoras theorem $r^2 = p^2 + q^2$. We evaluate $\tan^{-1}\left(\frac{p}{q+r}\right) + \tan^{-1}\left(\frac{q}{r+p}\right)$ using the addition formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$: Numerator: $\frac{p}{q+r} + \frac{q}{r+p} = \frac{p(r+p) + q(q+r)}{(q+r)(r+p)} = \frac{pr + p^2 + q^2 + qr}{(q+r)(r+p)}$. Denominator: $1 - \frac{pq}{(q+r)(r+p)} = \frac{(q+r)(r+p) - pq}{(q+r)(r+p)} = \frac{qr + pq + r^2 + pr - pq}{(q+r)(r+p)} = \frac{pr + r^2 + qr}{(q+r)(r+p)}$. Since $p^2 + q^2 = r^2$, the numerator becomes $pr + r^2 + qr$, which is identical to the denominator. Thus, the argument is $\frac{pr + r^2 + qr}{pr + r^2 + qr} = 1$. Therefore, $\tan^{-1}(1) = \frac{\pi}{4}$.