A) \[\frac{{{p}^{2}}+1}{{{p}^{2}}}\]
B) \[\frac{{{p}^{2}}+1}{\sqrt{p}}\]
C) \[\frac{{{p}^{2}}+1}{2p}\]
D) \[\frac{p+1}{2p}\]
Correct Answer: C
Solution :
We know, \[se{{c}^{2}}\theta -ta{{n}^{2}}\theta =1\] \[\Rightarrow \] \[(sec\theta +tan\theta )\text{ (}sec\theta -tan\theta )=1\] \[\Rightarrow \] \[\sec \theta -\tan \theta =\frac{1}{p}\] \[\therefore \]\[\sec \theta +\tan \theta +\sec \theta -\tan \theta =p+\frac{1}{p}\] \[\Rightarrow \] \[2\sec \theta =p+\frac{1}{p}\] \[\Rightarrow \] \[\sec \theta =\frac{{{p}^{2}}+1}{2p}\]
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