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

Revision as of 11:14, 14 April 2012

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 $\sec x + \tan x = \boxed{\frac{1}{2}}$ .

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

Revision as of 19:17, 5 June 2011 (view source) Minsoens (talk \| contribs) m (→‎Solution) ← Older edit		Revision as of 11:14, 14 April 2012 (view source) 1=2 (talk \| contribs) m Newer edit →
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		+	[[Category:Introductory Trigonometry Problems]]