1999 AMC 12 Problem 15

Attempt Problem 15 of the 1999 AMC 12 below, then check your answer against the professionally curated solution 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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15.

Let

x

be a real number such that

\sec x - \tan x = 2.

What is

\sec x + \tan x?

0.1

0.2

0.3

0.4

0.5

Recall the identity

\sec^2 x - \tan^2 x = 1

Factor it as

(\sec x - \tan x)(\sec x + \tan x)

Since

\sec^2 x - \tan^2 x = 1,

we have

\begin{aligned} &(\sec x - \tan x)(\sec x + \tan x) \\ &\quad = 1. \end{aligned}

With

\sec x - \tan x = 2,

it follows that

\sec x + \tan x = \tfrac12 = 0.5.

Thus, the correct answer is E.

Problem 15 in other years