A) \[\frac{({{e}^{x}}+{{e}^{-x}})}{2}\]
B) \[\frac{2}{({{e}^{x}}+{{e}^{-x}})}\]
C) \[\frac{({{e}^{x}}-{{e}^{-x}})}{2}\]
D) \[\frac{({{e}^{x}}-{{e}^{-x}})}{({{e}^{x}}+{{e}^{-x}})}\]
Correct Answer: B
Solution :
\[\tan \theta +\sec \theta ={{e}^{x}}\] ?..(i) \[\therefore \,\,\,\sec \theta -\tan \theta ={{e}^{-x}}\] ?..(ii) From (i) and (ii), \[\,2\sec \theta ={{e}^{x}}+{{e}^{-x}}\,\Rightarrow \,\cos \theta =\frac{2}{{{e}^{x}}+{{e}^{-x}}}.\]
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