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

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<math>(\sec x - \tan x)(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1</math>, so <math>\sec x + \tan x = \boxed{\frac{1}{2}}</math>.
 
<math>(\sec x - \tan x)(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1</math>, so <math>\sec x + \tan x = \boxed{\frac{1}{2}}</math>.
 
{{solution}}
 
  
 
==See Also==
 
==See Also==
  
 
{{AHSME box|year=1999|num-b=14|num-a=16}}
 
{{AHSME box|year=1999|num-b=14|num-a=16}}

Revision as of 19:17, 5 June 2011

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

$(\sec x - \tan x)(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1$, so $\sec x + \tan x = \boxed{\frac{1}{2}}$.

See Also

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