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 | + | 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 | ||
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==Solution== | ==Solution== | ||
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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{\textbf{(E)}\ 0.5}</math>. | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1999|num-b=14|num-a=16}} | {{AHSME box|year=1999|num-b=14|num-a=16}} | ||
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+ | [[Category:Introductory Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 11:41, 2 January 2016
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 |
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