If \[\tan \theta +\sec \theta ={{e}^{x}},\]then \[\cos \theta \] equals [AMU 2002]

\[\frac{({{e}^{x}}-{{e}^{-x}})}{2}\]

\[\frac{({{e}^{x}}+{{e}^{-x}})}{2}\]

\[\frac{2}{({{e}^{x}}+{{e}^{-x}})}\]

\[\frac{({{e}^{x}}-{{e}^{-x}})}{({{e}^{x}}+{{e}^{-x}})}\]

If \[\tan \theta +\sec \theta ={{e}^{x}},\]then \[\cos \theta \] equal

1.	If \[\tan \theta +\sec \theta ={{e}^{x}},\]then \[\cos \theta \] equals [AMU 2002]
A.	\[\frac{({{e}^{x}}+{{e}^{-x}})}{2}\]
B.	\[\frac{2}{({{e}^{x}}+{{e}^{-x}})}\]
C.	\[\frac{({{e}^{x}}-{{e}^{-x}})}{2}\]
D.	\[\frac{({{e}^{x}}-{{e}^{-x}})}{({{e}^{x}}+{{e}^{-x}})}\]
Answer» C. \[\frac{({{e}^{x}}-{{e}^{-x}})}{2}\]

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If \[\tan \theta +\sec \theta ={{e}^{x}},\]then \[\cos \theta \] equals [AMU 2002]

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