Difference between revisions of "1999 AHSME Problems/Problem 15"

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==Problem==
 
==Problem==
  
Let <math> x</math> be a real number such that <math> \sec x \minus{} \tan x = 2</math>. Then <math> \sec x \plus{} \tan x =</math>
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Let <math> x</math> be a real number such that <math> \sec x - \tan x = 2</math>. Then <math> \sec x + \tan x =</math>
  
 
<math> \textbf{(A)}\ 0.1 \qquad  
 
<math> \textbf{(A)}\ 0.1 \qquad  

Revision as of 20:40, 2 June 2011

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Problem

Let $x$ be a real number such that $\sec x - \tan x = 2$. Then $\sec x + \tan x =$

$\textbf{(A)}\ 0.1 \qquad  \textbf{(B)}\ 0.2 \qquad  \textbf{(C)}\ 0.3 \qquad  \textbf{(D)}\ 0.4 \qquad  \textbf{(E)}\ 0.5$

Solution

See Also

1999 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AHSME Problems and Solutions