1991 AIME Problems/Problem 9

Problem

Suppose that $\sec x+\tan x=\frac{22}7$ and that $\csc x+\cot x=\frac mn,$ where $\frac mn$ is in lowest terms. Find $m+n^{}_{}.$

Solution

Use the two trigonometric Pythagorean identities $\displaystyle 1 + \tan^2 x = \sec^2 x$ and $\displaystyle 1 + \cot^2 x = \csc^2 x$ .

If we square $\sec x = \frac{22}{7} - \tan x$ , we find that $\sec^2 x = (\frac{22}7)^2 - 2(\frac{22}7)\tan x + \tan^2 x$ , so $1 = (\frac{22}7)^2 - \frac{44}7 \tan x$ . Solving shows that $\tan x = \frac{435}{308}$ .

Call $y = \frac mn$ . Rewrite the second equation in a similar fashion: $\displaystyle 1 = y^2 - 2y\cot x$ . Substitute in $\cot x = \frac{1}{\tan x} = \frac{308}{435}$ to get a quadratic: $0 = y^2 - \frac{616}{435} - 1$ . The quadratic is factorable (though somewhat ugly); $(15y - 29)(29y + 15) = 0$ . It turns out that only the positive root will work, so the value of $y = \frac{29}{15}$ and $m + n = 044$ .

1991 AIME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1991 AIME Problems/Problem 9

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