A) \[\frac{ac}{{{a}^{2}}+{{c}^{2}}}\]
B) \[\frac{2ac}{a-c}\]
C) \[\frac{2ac}{{{a}^{2}}-{{c}^{2}}}\]
D) \[\frac{a+c}{1-ac}\]
Correct Answer: C
Solution :
[c] Let \[{{\tan }^{-1}}x=\alpha \] and \[{{\tan }^{-1}}y=\beta \Rightarrow \tan \alpha =x,\tan \beta =y\] The given system of equations is \[a\tan \alpha +b\sec \alpha =c\] and \[a\tan \beta +b\sec \beta =c\] \[\therefore \,\,\,\alpha \,\,and\,\,\beta \] are the roots of \[a\tan \theta +b\sec \theta =c\] \[\Rightarrow {{(bsec\theta )}^{2}}={{(c-atan\theta )}^{2}}\] \[\Rightarrow ({{a}^{2}}-{{b}^{2}})ta{{n}^{2}}\theta -2ac\tan \theta +{{c}^{2}}-{{b}^{2}}=0\] \[\Rightarrow \tan \alpha +tan\beta =\frac{2ac}{{{a}^{2}}-{{b}^{2}}}\] And \[\tan \alpha \tan \beta =\frac{{{c}^{2}}-{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}\] \[\Rightarrow x+y=\frac{2ac}{{{a}^{2}}-{{b}^{2}}}\] and \[xy=\frac{{{c}^{2}}-{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}\] \[\Rightarrow 1-xy=\frac{{{a}^{2}}-{{c}^{2}}}{{{a}^{2}}-{{b}^{2}}}\Rightarrow \frac{x+y}{1-xy}=\frac{2ac}{{{a}^{2}}-{{c}^{2}}}\]
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