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>. | ||
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==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 be a real number such that . Then
Solution
, so .
See Also
1999 AHSME (Problems • Answer Key • Resources) | ||
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 |