A) \[\frac{b}{1+a}\].
B) \[\frac{-b}{1+a}\]
C) \[\frac{a}{1+b}\]
D) None of these
Correct Answer: B
Solution :
Here the equation is \[a{{x}^{2}}-bxy-{{y}^{2}}=0\] and given that \[{{m}_{1}}=\tan \alpha \] and \[{{m}_{2}}=\tan \beta \]and we know that \[{{m}_{1}}+{{m}_{2}}=\frac{b}{-1}=\tan \alpha +\tan \beta \] and \[{{m}_{1}}{{m}_{2}}=\frac{a}{-1}=\tan \alpha .\tan \beta \] \[\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{-b}{1-(-a)}=\frac{-b}{(1+a)}\].
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