question_answer

AB is a vertical pole. The end A is on the ground, C is the middle point of AB, P is a point on the level ground. The portion BC subtends an angle \[\alpha \] at P. If \[AP=n\cdot AB,\] then \[\tan \alpha \] is equal to

A) \[\frac{n}{2{{n}^{2}}+1}\]     

B) \[\frac{n}{{{n}^{2}}-1}\]

C) \[\frac{n}{{{n}^{2}}+1}\]                  

D) \[\frac{{{n}^{2}}-1}{{{n}^{2}}+1}\]

Correct Answer: A

Solution :

Let \[\angle CPA=\beta ,\]

\[\tan \beta =\frac{AC}{AP}=\frac{\frac{1}{2}AB}{n\,\,AB}=\frac{1}{2n}\]

and \[\tan (\alpha +\beta )=\frac{AB}{AP}=\frac{AB}{nAB}=\frac{1}{n}\]

\[\Rightarrow \]   \[\frac{\tan \alpha +\tan \beta }{1-\tan \alpha tan\beta }=\frac{1}{n}\]

\[\Rightarrow \]   \[n\,(\tan \alpha +\tan \beta )=1-\tan \alpha \tan \beta \]

\[\Rightarrow \]   \[n\,tan\alpha +\tan \alpha \tan \beta =1-n\tan \beta \]

\[\Rightarrow \]   \[\tan \alpha =\frac{1-n\,\,\tan \beta }{n+\tan \beta }=\frac{1-\frac{1}{2}}{n+\frac{1}{2n}}=\frac{n}{2{{n}^{2}}+1}\]