A) \[\frac{{{l}^{2}}-1}{2l}\]
B) \[\frac{{{l}^{2}}+1}{2l}\]
C) \[\frac{{{l}^{2}}-1}{l}\]
D) \[\frac{2\left( {{l}^{2}}-1 \right)}{l}\]
Correct Answer: B
Solution :
Given, \[\tan \theta +\sec \theta =I\] ...(i) \[\Rightarrow \frac{\left( \tan \theta +\sec \theta \right)\left( \sec \theta -\tan \theta \right)}{\left( \sec \theta -\tan \theta \right)}=I\] \[\Rightarrow \,\,\,\frac{\left( {{\sec }^{2}}\theta -{{\tan }^{2}}\theta \right)}{\left( \sec \theta -tan\theta \right)}=I\] \[\Rightarrow \,\,\frac{1}{\sec \theta -\tan \theta }=I\] \[\left[ \because \,\,{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \right]\] \[\Rightarrow \,\,\sec \theta -\tan \theta =\frac{1}{I}\] (ii) On adding Eqs. (i) and (ii), we get \[2\sec \theta =I+\frac{1}{I}\] \[\Rightarrow \,\,\sec \theta =\frac{{{I}^{2}}+1}{2I}\]
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