1999 AMC 12 Problem 15

Below is the professionally curated solution for Problem 15 of the 1999 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 12 solutions, or check the answer key.

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Concepts:trigonometric identity difference of squares

Difficulty rating: 1550

15.

Let $x$ be a real number such that $\sec x - \tan x = 2.$ Then $\sec x + \tan x = \, ?$

$0.1$

$0.2$

$0.3$

$0.4$

$0.5$

Solution:

Since $\sec^2 x - \tan^2 x = 1,$ we have $(\sec x - \tan x)(\sec x + \tan x) = 1.$ With $\sec x - \tan x = 2,$ it follows that $\sec x + \tan x = \tfrac12 = 0.5.$

Thus, the correct answer is E.

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