A) \[\frac{{{m}^{2}}-1}{2m}\]
B) \[\frac{{{m}^{2}}+1}{2m}\]
C) \[\frac{m+1}{m}\]
D) \[\frac{{{m}^{2}}+1}{m}\]
Correct Answer: B
Solution :
\[\tan \theta +\sec \theta =m\] \[\Rightarrow \,\,\sec \theta =m-\tan \theta \] On squaring both sides, we get \[{{\left( \sec \theta \right)}^{2}}={{\left( m-\tan \theta \right)}^{2}}\] \[\Rightarrow \,\,{{\sec }^{2}}\theta ={{m}^{2}}+{{\tan }^{2}}\theta -2m\,\tan \theta \] \[\Rightarrow \,{{\sec }^{2}}\theta -{{\tan }^{2}}\theta ={{m}^{2}}-2m\,\tan \theta \] \[\Rightarrow \,\,1={{m}^{2}}-2m\,\tan \theta \] \[(\because \,\,\,{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1)\] \[\Rightarrow \,\,\tan \theta =\frac{{{m}^{2}}-1}{2m}\] On putting the value of \[tan\text{ }\theta \]in initial equation, we get \[\frac{{{m}^{2}}-1}{2m}+\sec \theta =m\Rightarrow \sec \theta =m-\frac{{{m}^{2}}-1}{2m}\] \[\therefore \,\,\sec \theta =\frac{2{{m}^{2}}-{{m}^{2}}+1}{2m}=\]
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