1999 AHSME Problems/Problem 15

Revision as of 18:26, 1 May 2023 by Songmath20 (talk | contribs) (Solution 2 (Alternate))

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 $(1-\sin^{2}x)/\cos^{2}x = 1$. This equates to $2y = 1$. Thus, $\boxed{\textbf{(E)}\ 0.5}$.

See Also

1999 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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