A) \[2tan\text{ }\beta tan\,\gamma \]
B) \[tan\text{ }\beta +2\text{ }tan\text{ }\gamma \]
C) \[tan\text{ }\beta -2\text{ }tan\text{ }\gamma \]
D) \[2\left( tan\text{ }\beta +tan\text{ }\gamma \text{ } \right)\]
Correct Answer: B
Solution :
[b] \[\because \alpha +\beta =\frac{\pi }{2}\Rightarrow \alpha =\frac{\pi }{2}-\beta \] \[\Rightarrow tan\alpha =tan\left( \frac{\pi }{2}-\beta \right)=cot\beta =\frac{1}{\tan \beta }\] \[\Rightarrow tan\alpha \,tan\beta =1\] ............(i) \[\therefore \alpha =\beta +\gamma \] \[tan\alpha =tan\left( \beta +\gamma \right)=\frac{tan\beta +\tan \gamma }{1-\tan \beta .tan\gamma }\] \[\Rightarrow \] \[tan\beta +tan\gamma =tan\alpha -tan\alpha .tan\beta .\tan \gamma \] \[=tan\alpha -1.tan\beta \left[ from\left( i \right) \right]\] \[\Rightarrow \] \[tan\beta +2.\tan \gamma =tan\,\alpha \]
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