1999 AHSME Problems/Problem 15

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, $\boxed{\textbf{(E)}\ 0.5}$ .

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1999 AHSME Problems/Problem 15

Contents

Problem

Solution 1 (Fastest)

Solution 2 (Alternate)

See Also