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

Revision as of 19:30, 1 May 2023

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 1 (Fastest)

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

Solution 2 (Alternate)

Note that $\sec x - \tan x = (1-\sin x)/\cos x$ , and $\sec x + \tan x = (1+\sin x)/\cos x$ . Let $(1+\sin x)/\cos x = y$ . Multiplying the two, we get $(1-\sin^{2}x)/\cos^{2}x = 1$ .Then, $2y = 1$ . $\sec x + \tan x = \boxed{\textbf{(E)}\ 0.5}$ .

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 14: / Line 14: @@
 ==Solution 2 (Alternate)==
-Note that <math>\sec x - \tan x = (1-\sin x)/\cos x</math>, and <math>\sec x + \tan x = (1+\sin x)/\cos x</math>, let <math>(1+\sin x)/\cos x = y</math>. Multiplying the two, we get <math>(1-\sin^{2}x)/\cos^{2}x = 1</math>.Then, <math>2y = 1</math>. <math>\sec x + \tan x =
+Note that <math>\sec x - \tan x = (1-\sin x)/\cos x</math>, and <math>\sec x + \tan x = (1+\sin x)/\cos x</math>. Let <math>(1+\sin x)/\cos x = y</math>. Multiplying the two, we get <math>(1-\sin^{2}x)/\cos^{2}x = 1</math>.Then, <math>2y = 1</math>. <math>\sec x + \tan x =
 \boxed{\textbf{(E)}\ 0.5}</math>.

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

Revision as of 19:30, 1 May 2023

Contents

Problem

Solution 1 (Fastest)

Solution 2 (Alternate)

See Also