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

m (Solution)
Line 12: Line 12:
  
 
<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 20:17, 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions