A) \[9{{b}^{2}}\]
B) \[\frac{9}{{{a}^{2}}}\]
C) \[\frac{-2}{b}\]
D) 2
Correct Answer: A
Solution :
\[a\sec \theta +b\tan \theta =1\] \[\Rightarrow \] \[a\,\sec \theta =1-b\tan \theta \] ?.(i) \[\Rightarrow \] \[{{(a\,sec\theta )}^{2}}={{(1-b\tan \theta )}^{2}}\] Also,\[{{a}^{2}}\,{{\sec }^{2}}\theta -{{b}^{2}}{{\tan }^{2}}\theta =5\] (Given) ?.(ii) \[\therefore \] Substituting \[{{a}^{2}}{{\sec }^{2}}\theta ={{(1-b\,tan\theta )}^{2}}\]in (ii) i.e., \[{{(1-b\tan \theta )}^{2}}-{{b}^{2}}{{\tan }^{2}}\theta =5\] \[\Rightarrow \] \[1-2b\tan \theta =5\] \[\Rightarrow \] \[\tan \theta =-\frac{4}{2b}=\frac{-2}{b}\] ??(iii) Using (iii) in (i), we get \[\sec \theta =\frac{1-b\left( -\frac{2}{b} \right)}{a}=\frac{3}{a}\] and \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \] \[\Rightarrow \] \[1+\frac{4}{{{b}^{2}}}=\frac{9}{{{a}^{2}}}\Rightarrow \,{{a}^{2}}{{b}^{2}}+4{{a}^{2}}=9{{b}^{2}}\]
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