A) \[-2a,\frac{1}{2a}\]
B) \[-\frac{1}{2a},2a\]
C) \[2a\]
D) \[\frac{1}{2a},2a\]
Correct Answer: A
Solution :
Let \[\sec \theta -\tan \theta =\lambda \] ...(i) Also, \[\sec \theta +\tan \theta =\frac{1}{\lambda }\] ...(ii) On subtracting Eq. (i) from Eq. (ii), we get \[2\tan \theta =\frac{1}{\lambda }-\lambda \] \[\Rightarrow \] \[2\left( a-\frac{1}{4a} \right)=\frac{1}{\lambda }-\lambda \] \[\Rightarrow \] \[2a-\frac{1}{2a}=\frac{1}{\lambda }-\lambda \] \[\therefore \] \[\lambda =\frac{1}{2a},-2a\]
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