A) \[n=({{n}^{2}}+1)\tan \alpha \]
B) \[n=(2{{n}^{2}}-1)\tan \alpha \]
C) \[{{n}^{2}}=(2{{n}^{2}}+1)\tan \alpha \]
D) \[n=(2{{n}^{2}}+1)\tan \alpha \]
Correct Answer: D
Solution :
\[\tan \alpha =\frac{-\frac{AC}{AP}+\frac{AB}{AP}}{1+\frac{AC}{AP}.\frac{AB}{AP}}\] \[\{AP=n(AB)\] \[\Rightarrow AP=2n(AC)\}\] \[\tan \alpha =\frac{-\frac{1}{2n}+\frac{1}{n}}{1+\frac{1}{2{{n}^{2}}}}\] \[\Rightarrow \] \[\frac{n}{(2{{n}^{2}}+1)}=\tan \alpha \]\[\Rightarrow \] \[n=(2{{n}^{2}}+1)\tan \alpha \].You need to login to perform this action.
You will be redirected in
3 sec