(d) \(\frac{n}{2n^2+1}\)
Let the height of the vertical pole AB = h metres.
Then, AC = BC = \(\frac{h}{2}\), AP = n. AB = \(\frac{n}{h}\)
Let ∠APC = α.
Given ∠BPC = β
∴ ∠APB = α + β
In rt. Δ ACP,
tan α = \(\frac{AC}{AP}\) = \(\frac{\frac{h}{2}}{n\,h}\) = \(\frac{1}{2h}\) ....(i)
In rt. Δ BAP, tan (α + β) = \(\frac{AB}{AP}\) = \(\frac{h}{n\,h}\) = \(\frac{1}{n}\) .....(ii)
Now tan b = tan {(α + β) – α}
= \(\frac{\text{tan}\,(\alpha+\beta)-\text{tan}\,\alpha}{1+\text{tan}\,(\alpha+\beta)\,\text{tan}\,\alpha}\)
= \(\frac{\frac{1}{n}-\frac{1}{2n}}{1+\frac{1}{n}.\frac{1}{2n}}\) = \(\frac{n}{2n^2+1}\). (Using (i) and (ii))