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

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==Solution 2 (Alternate)==
 
==Solution 2 (Alternate)==
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Note that <math>\sec x - \tan x = (1-\sin x)/\cos x</math>, and we can say that <math>\sec x + \tan x = (1+\sin x)/\cos x</math>, and if we let that equal y and multiply the two, we get <math>(1-\sin^{2}x)/\cos^{2}x</math>, which equals <math>1</math>. This equates to <math>2y = 1</math>. Thus, \boxed{\textbf{(E)}\ 0.5}$.
+
Note that <math>\sec x - \tan x = (1-\sin x)/\cos x</math>, and since <math>\sec x + \tan x = (1+\sin x)/\cos x</math>, let <math>(1+\sin x)/\cos x = y</math> and multiply the two, we get <math>(1-\sin^{2}x)/\cos^{2}x</math>, which equals <math>1</math>. This equates to <math>2y = 1</math>. Thus, \boxed{\textbf{(E)}\ 0.5}$.
  
 
==See Also==
 
==See Also==

Revision as of 19:24, 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 since $\sec x + \tan x = (1+\sin x)/\cos x$, let $(1+\sin x)/\cos x = y$ and multiply the two, we get $(1-\sin^{2}x)/\cos^{2}x$, which equals $1$. This equates to $2y = 1$. Thus, \boxed{\textbf{(E)}\ 0.5}$.

See Also

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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