1999 AHSME Problems/Problem 15: Difference between revisions

Revision as of 10:40, 2 January 2016

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

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

$(\sec x - \tan x)(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1$ , so $\sec x + \tan x = \boxed{(E)\ 0.5}$ .

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AHSME Problems and Solutions

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 ==Solution==
-<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{(E)\ 0.5}</math>.
 ==See Also==