A) \[\frac{1}{2}\]
B) \[\frac{2}{3}\]
C) 1
D) 2
Correct Answer: A
Solution :
| [a] \[\tan \beta =\frac{h}{CD}\]\[\Rightarrow \]\[CD=\frac{h}{\tan \beta }=h\cot \beta \] and \[\tan \alpha =\frac{h}{\frac{h}{2}+CD}\] |
 |
| \[\Rightarrow \] \[\frac{h}{2}+CD=\frac{h}{\tan \alpha }=h\cot \alpha \] |
| \[\Rightarrow \] \[\frac{h}{2}+h\,\,\cot \beta =h\,\,\cot \alpha \] |
| \[\Rightarrow \]\[h\cot \alpha -h\cos \beta =\frac{h}{2}\]\[\Rightarrow \]\[(\cot \alpha -\cot \beta )=\frac{h}{2}\] |
| \[\therefore \] \[\cot \alpha -cot\beta =\frac{1}{2}\] |
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