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

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==Problem==
 
==Problem==
  
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\textbf{(E)}\ 0.5</math>
 
\textbf{(E)}\ 0.5</math>
  
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==Solution==
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==Solution 1 (Fastest)==
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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>.
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==Solution 2 (Alternate, Slightly Longer)==
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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, we get <math>(1-\sin^{2}x)/\cos^{2}x = 1</math>.Then, <math>2y = 1</math>. <math>\sec x + \tan x =
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\boxed{\textbf{(E)}\ 0.5}</math>. ~songmath20 
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Edited 5.1.2023
  
 
==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]]
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{{MAA Notice}}

Latest revision as of 18:36, 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, Slightly Longer)

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, we get $(1-\sin^{2}x)/\cos^{2}x = 1$.Then, $2y = 1$. $\sec x + \tan x = \boxed{\textbf{(E)}\ 0.5}$. ~songmath20 Edited 5.1.2023

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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