\[\sec {{50}^{o}}+\tan {{50}^{o}}\] is equal to [DCE 2002]

1.	\[\sec {{50}^{o}}+\tan {{50}^{o}}\] is equal to [DCE 2002]
A.	\[\tan {{20}^{o}}+\tan {{50}^{o}}\]
B.	\[2\tan {{20}^{o}}+\tan {{50}^{o}}\]
C.	\[\tan {{20}^{o}}+2\tan {{50}^{o}}\]
D.	\[2\tan {{20}^{o}}+2\tan {{50}^{o}}\]
Answer» D. \[2\tan {{20}^{o}}+2\tan {{50}^{o}}\]

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\[\sec {{50}^{o}}+\tan {{50}^{o}}\] is equal to [DCE 2002]

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