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

Solution 1

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

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

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

Solution 2

Make the substitution $u = \tan \frac x2$ (a substitution commonly used in calculus). $\tan \frac x2 = \frac{\sin x}{1+\cos x}$ , so $\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn$ . $\sec x + \tan x = \frac{1 + \sin x}{\cos x}.$ Now note the following: $\begin{align*}\sin x &= \frac{2u}{1+u^2}\\ \cos x &= \frac{1-u^2}{1+u^2}\end{align*}$ Plugging these into our equality gives: $\frac{1+\frac{2u}{1+u^2}}{\frac{1-u^2}{1+u^2}} = \frac{22}7$

This simplifies to $\frac{1+u}{1-u} = \frac{22}7$ , and solving for $u$ gives $u = \frac{15}{29}$ , and $\frac mn = \frac{29}{15}$ . Finally, $m+n = 044$ .

1991 AIME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1991 AIME Problems/Problem 9

Problem

Contents

Solution

Solution 1

Solution 2

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