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 the given $\sec x = \frac{22}{7} - \tan x$ , we find that

$\begin{align*} \sec^2 x &= \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x \\ 1 &= \left(\frac{22}7\right)^2 - \frac{44}7 \tan x \end{align*}$

This yields $\tan x = \frac{435}{308}$ .

Let $y = \frac mn$ . Then squaring,

$\csc^2 x = (y - \cot x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.$

Substituting $\cot x = \frac{1}{\tan x} = \frac{308}{435}$ yields a quadratic equation: $0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)$ . 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

Recall that $\sec^2 x - \tan^2 x = 1$ , from which we find that $\sec x - \tan x = 7/22$ . Adding the equations

$\begin{eqnarray*} \sec x + \tan x & = & 22/7 \\ \sec x - \tan x & = & 7/22\end{eqnarray*}$

together and dividing by 2 gives $\sec x = 533/308$ , and subtracting the equations and dividing by 2 gives $\tan x = 435/308$ . Hence, $\cos x = 308/533$ and $\sin x = \tan x \cos x = (435/308)(308/533) = 435/533$ . Thus, $\csc x = 533/435$ and $\cot x = 308/435$ . Finally,

$\csc x + \cot x = \frac {841}{435} = \frac {29}{15},$

so $m + n = 044$ .

Solution 3 (least computation)

By the given, $\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}$ and $\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k$ .

Multiplying the two, we have

$\frac {1}{\sin x \cos x} + \frac {1}{\sin x} + \frac {1}{\cos x} + 1 = \frac {22}{7}k$

Subtracting both of the two given equations from this, and simpliyfing with the identity $\frac {\sin x}{\cos x} + \frac {\cos x}{\sin x} = \frac{\sin ^2 x + \cos ^2 x}{\sin x \cos x} = \frac {1}{\sin x \cos x}$ , we get

$1 = \frac {22}{7}k - \frac {22}{7} - k.$

Solving yields $k = \frac {29}{15}$ , and $m+n = 044$

Solution 4

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, $m+n = 044$ .

Solution 5

We are given that $\frac{1+\sin x}{\cos x}=\frac{22}7\implies\frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}$ $=\frac{\cos x}{1-\sin x}$ , or equivalently, $\cos x=\frac{7+7\sin x}{22}=\frac{22-22\sin x}7\implies\sin x=\frac{22^2-7^2}{22^2+7^2}$ $\implies\cos x=\frac{2\cdot22\cdot7}{22^2+7^2}$ . Note that what we want is just $\frac{1+\cos x}{\sin x}=\frac{1+\frac{2\cdot22\cdot7}{22^2+7^2}}{\frac{22^2-7^2}{22^2+7^2}}=\frac{22^2+7^2+2\cdot22\cdot7}{22^2-7^2}=\frac{(22+7)^2}{(22-7)(22+7)}=\frac{22+7}{22-7}$ $=\frac{29}{15}\implies m+n=29+15=\boxed{044}$ .

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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