Difference between revisions of "1991 AIME Problems/Problem 9"

Latest revision as of 13:04, 20 January 2023

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 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 $m + n = \boxed{044}$ .

Note: The problem is much easier computed if we consider what $\sec (x)$ is, then find the relationship between $\sin( x)$ and $cos (x)$ (using $\tan (x) = \frac{435}{308}$ , and then computing $\csc x + \cot x$ using $1/\sin x$ and then the reciprocal of $\tan x$ .

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). By the half-angle identity for tangent, $\tan \frac x2 = \frac{\sin x}{1+\cos x}$ , so $\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn$ . Also, we have $\sec x + \tan x = \frac{1 + \sin x}{\cos x}.$ 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 $\frac{1+\cos x}{\sin x}=\frac{1+\frac{2\cdot22\cdot7}{22^2+7^2}}{\frac{22^2-7^2}{22^2+7^2}}=\frac{22^2+7^2+2\cdot22\cdot7}{22^2-7^2}=\frac{(22+7)^2}{(22-7)(22+7)}=\frac{22+7}{22-7}$ $=\frac{29}{15}\implies m+n=29+15=\boxed{044}$ .

Solution 6

Assign a right triangle with angle $x$ , hypotenuse $c$ , adjacent side $a$ , and opposite side $b$ . Then, through the given information above, we have that..

$\frac{c}{a}+\frac{b}{a}=\frac{22}{7}\implies \frac{c+b}{a}=\frac{22}{7}$

$\frac{c}{b}+\frac{a}{b}=\frac{m}{n}\implies \frac{a+c}{b}=\frac{m}{n}$

Hence, because similar right triangles can be scaled up by a factor, we can assume that this particular right triangle is indeed in simplest terms.

Hence, $a=7$ , $b+c=22$

Furthermore, by the Pythagorean Theorem, we have that

$a^2+b^2=c^2\implies 49+b^2=c^2$

Solving for $c$ in the first equation and plugging in into the second equation...

$49+b^2=(22-b)^2\implies 49+b^2=484-44b+b^2\implies 44b=435\implies b=\frac{435}{44}$

Hence, $c=22-\frac{435}{44}=\frac{533}{44}$

Now, we want $\frac{a+c}{b}$

Plugging in, we find the answer is $\frac{\frac{7\cdot{44}}{44}+\frac{533}{44}}{\frac{435}{44}}=\frac{841}{435}=\frac{29}{15}$

Hence, the answer is $29+15=\boxed{044}$

Solution 7

We know that $\sec(x) = \frac{h}{a}$ and that $\tan(x) = \frac{o}{a}$ where $h$ , $a$ , $o$ represent the hypotenuse, adjacent, and opposite (respectively) to angle $x$ in a right triangle. Thus we have that $\sec(x) + \tan(x) = \frac{h+o}{a}$ . We also have that $\csc(x) + \cot(x) = \frac{h}{o} + \frac{a}{o} = \frac{h+a}{o}$ . Set $\sec(x) + \tan(x) = \alpha$ and csc(x)+cot(x) = $\beta$ . Then, notice that $\alpha + \beta = \frac{h+o}{a} + \frac{h+a}{o} = \frac{oh+ah+o^2 + a^2}{oa} = \frac{h(o+a+h)}{oa}$ ( This is because of the Pythagorean Theorem, recall $o^2 +a^2 = h^2$ ). But then notice that $\alpha \cdot \beta = \frac{(o+h)(a+h)}{oa} = \frac{oa +oh +ha +h^2}{oa} = 1+ \frac{h(o+a+h)}{oa} = 1+ \alpha + \beta$ . From the information provided in the question, we can substitute $\alpha$ for $\frac{22}{7}$ . Thus, $\frac{22 \beta}{7}= \beta + \frac{29}{7} \Longrightarrow 22 \beta = 7 \beta + 29 \Longrightarrow 15 \beta = 29 \Longrightarrow \beta = \frac{29}{15}$ . Since, essentially we are asked to find the sum of the numerator and denominator of $\beta$ , we have $29 + 15 = \boxed{044}$ .

~qwertysri987

Solution 8

Firstly, we write $\sec x+\tan x=a/b$ where $a=22$ and $b=7$ . This will allow us to spot factorable expressions later. Now, since $\sec^2x-\tan^2x=1$ , this gives us $\sec x-\tan x=\frac{b}{a}$ Adding this to our original expressions gives us $2\sec x=\frac{a^2+b^2}{ab}$ or $\cos x=\frac{2ab}{a^2+b^2}$ Now since $\sin^2x+\cos^2x=1$ , $\sin x=\sqrt{1-\cos^2x}$ So we can write $\sin x=\sqrt{1-\frac{4a^2b^2}{(a^2+b^2)^2}}$ Upon simplification, we get $\sin x=\frac{a^2-b^2}{a^2+b^2}$ We are asked to find $1/\sin x+\cos x/\sin x$ so we can write that as $\csc x+\cot x=\frac{1}{\sin x}+\frac{\cos x}{\sin x}$ $\csc x+\cot x=\frac{a^2+b^2}{a^2-b^2}+\frac{2ab}{a^2+b^2}\frac{a^2+b^2}{a^2-b^2}$ $\csc x+\cot x=\frac{a^2+b^2+2ab}{a^2-b^2}$ $\csc x+\cot x=\frac{(a+b)^2}{(a-b)(a+b)}$ $\csc x+\cot x=\frac{a+b}{a-b}$ Now using the fact that $a=22$ and $b=7$ yields, $\csc x+\cot x=\frac{29}{15}=\frac{p}{q}$ so $p+q=15+29=\boxed{44}$

~Chessmaster20000

Solution 9

Rewriting $\sec{x}$ and $\tan{x}$ in terms of $\sin{x}$ and $\cos{x}$ , we know that $\frac{1+\sin{x}}{\cos{x}}=\frac{22}{7}.$

Clearing fractions, $22\cos{x}=7+7\sin{x}.$

Squaring to get an expression in terms of $\sin^2{x}$ and $\cos^2{x}$ , $484\cos^2{x}=49+49\sin^2{x}+98\sin{x}.$

Substituting $\cos^2{x}=1-\sin^2{x},$

$484(1-\sin^2{x})=49+49\sin^2{x}+98\sin{x}.$

Expanding then collecting terms yields a quadratic in $\sin{x}:$

$533\sin^2{x}+98\sin{x}-435=0.$

To make calculations easier, let $y=\sin{x}.$

$533y^2+98y-435=0.$

Upon inspection, $y=-1$ is a root. Dividing by $y+1$ ,

$533y^2+98y-435=(533y-435)(y+1).$

Substituting $y=\sin{x},$ we see that $\sin{x}=-1$ doesn't work, as $\cos{x}=0$ , leaving $\tan{x}$ undefined.

We conclude that $\sin{x}=\frac{435}{533}.$

Since $\sin^2{x}+\cos^2{x}=1,$

$\cos{x}=\pm \sqrt{\frac{533^2-435^2}{533^2}}.$ $=\pm \frac{308}{533}.$

After checking via the given equation, we know that only the positive solution works.

Therefore,

$\csc{x}+\cot{x}=\frac{1}{\sin{x}}+\frac{\cos{x}}{\sin{x}}$ $=\frac{533}{435}+\frac{308}{435}$ $=\frac{29}{15}=\frac{m}{n}.$

Adding $m$ and $n$ , our answer is $\boxed{\textbf{044}}.$

-Benedict T (countmath1)

@@ Line 1: / Line 1: @@
+__TOC__
 == Problem ==
 Suppose that <math>\sec x+\tan x=\frac{22}7</math> and that <math>\csc x+\cot x=\frac mn,</math> where <math>\frac mn</math> is in lowest terms.  Find <math>m+n^{}_{}.</math>
-== Solution ==
+== Solution 1 ==
-Use the two [[Trigonometric identities#Pythagorean Identities|trigonometric Pythagorean identities]] <math>\displaystyle 1 + \tan^2 x = \sec^2 x</math> and <math>\displaystyle 1 + \cot^2 x = \csc^2 x</math>.
+Use the two [[Trigonometric identities#Pythagorean Identities|trigonometric Pythagorean identities]] <math>1 + \tan^2 x = \sec^2 x</math> and <math>1 + \cot^2 x = \csc^2 x</math>.
+If we square the given <math>\sec x = \frac{22}{7} - \tan x</math>, we find that
+<cmath>\begin{align*}
+\sec^2 x &= \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x \\
+&= \left(\frac{22}7\right)^2 - \frac{44}7 \tan x \end{align*}</cmath>
+This yields <math>\tan x = \frac{435}{308}</math>.
+Let <math>y = \frac mn</math>. Then squaring,
+<cmath>\csc^2 x = (y - \cot x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.</cmath>
+Substituting <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> yields a [[quadratic equation]]: <math>0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)</math>. It turns out that only the [[positive]] root will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = \boxed{044}</math>.
+Note: The problem is much easier computed if we consider what <math>\sec (x)</math> is, then find the relationship between <math>\sin( x)</math> and <math>cos (x)</math> (using <math>\tan (x) = \frac{435}{308}</math>, and then computing <math>\csc x + \cot x</math> using <math>1/\sin x</math> and then the reciprocal of <math>\tan x</math>.
+== Solution 2 ==
+Recall that <math>\sec^2 x - \tan^2 x = 1</math>, from which we find that <math>\sec x - \tan x = 7/22</math>. Adding the equations
+<cmath>\begin{eqnarray*} \sec x + \tan x & = & 22/7 \\
+\sec x - \tan x & = & 7/22\end{eqnarray*}</cmath>
+together and dividing by 2 gives <math>\sec x = 533/308</math>, and subtracting the equations and dividing by 2 gives <math>\tan x = 435/308</math>. Hence, <math>\cos x = 308/533</math> and <math>\sin x = \tan x \cos x = (435/308)(308/533) = 435/533</math>. Thus, <math>\csc x = 533/435</math> and <math>\cot x = 308/435</math>. Finally,
+<cmath>\csc x + \cot x = \frac {841}{435} = \frac {29}{15},</cmath>
+so <math>m + n = 044</math>.
+== Solution 3 (least computation)==
+By the given,
+<math>\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}</math> and
+<math>\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k</math>.
+Multiplying the two, we have
+<cmath>\frac {1}{\sin x \cos x} + \frac {1}{\sin x} + \frac {1}{\cos x} + 1 = \frac {22}{7}k</cmath>
+Subtracting both of the two given equations from this, and simpliyfing with the identity <math>\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}</math>, we get
+<cmath>1 = \frac {22}{7}k - \frac {22}{7} - k.</cmath>
+Solving yields <math>k = \frac {29}{15}</math>, and <math>m+n = 044</math>
+== Solution 4 ==
+Make the substitution <math>u = \tan \frac x2</math> (a substitution commonly used in calculus). By the half-angle identity for tangent, <math>\tan \frac x2 = \frac{\sin x}{1+\cos x}</math>, so <math>\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn</math>. Also, we have <math>\sec x + \tan x = \frac{1 + \sin x}{\cos x}.</math> Now note the following:
+<cmath>\begin{align*}\sin x &= \frac{2u}{1+u^2}\\
+\cos x &= \frac{1-u^2}{1+u^2}\end{align*}</cmath>
+Plugging these into our equality gives:
+<cmath>\frac{1+\frac{2u}{1+u^2}}{\frac{1-u^2}{1+u^2}} = \frac{22}7</cmath>
+This simplifies to <math>\frac{1+u}{1-u} = \frac{22}7</math>, and solving for <math>u</math> gives <math>u = \frac{15}{29}</math>, and <math>\frac mn = \frac{29}{15}</math>. Finally, <math>m+n = 044</math>.
+== Solution 5 ==
+We are given that <math>\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)}</math>
+<math>=\frac{\cos x}{1-\sin x}</math>, or equivalently, <math>\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}</math>
+<math>\implies\cos x=\frac{2\cdot22\cdot7}{22^2+7^2}</math>. Note that what we want is just <math>\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}</math>
+<math>=\frac{29}{15}\implies m+n=29+15=\boxed{044}</math>.
+== Solution 6 ==
+Assign a right triangle with angle <math>x</math>, hypotenuse <math>c</math>, adjacent side <math>a</math>, and opposite side <math>b</math>.
+Then, through the given information above, we have that..
+<math>\frac{c}{a}+\frac{b}{a}=\frac{22}{7}\implies \frac{c+b}{a}=\frac{22}{7}</math>
+<math>\frac{c}{b}+\frac{a}{b}=\frac{m}{n}\implies \frac{a+c}{b}=\frac{m}{n}</math>
+Hence, because similar right triangles can be scaled up by a factor, we can assume that this particular right triangle is indeed in simplest terms.
+Hence, <math>a=7</math>, <math>b+c=22</math>
+Furthermore, by the Pythagorean Theorem, we have that
+<math>a^2+b^2=c^2\implies 49+b^2=c^2</math>
+Solving for <math>c</math> in the first equation and plugging in into the second equation...
+<math>49+b^2=(22-b)^2\implies 49+b^2=484-44b+b^2\implies 44b=435\implies b=\frac{435}{44}</math>
+Hence, <math>c=22-\frac{435}{44}=\frac{533}{44}</math>
+Now, we want <math>\frac{a+c}{b}</math>
+Plugging in, we find the answer is <math>\frac{\frac{7\cdot{44}}{44}+\frac{533}{44}}{\frac{435}{44}}=\frac{841}{435}=\frac{29}{15}</math>
+Hence, the answer is <math>29+15=\boxed{044}</math>
+== Solution 7 ==
+We know that <math>\sec(x) = \frac{h}{a} </math> and that <math>\tan(x) = \frac{o}{a} </math> where <math>h</math>, <math>a</math>, <math>o</math> represent the hypotenuse, adjacent, and opposite (respectively) to angle <math>x</math> in a right triangle. Thus we have that <math>\sec(x) + \tan(x) = \frac{h+o}{a}</math>. We also have that <math>\csc(x) + \cot(x) = \frac{h}{o} + \frac{a}{o} = \frac{h+a}{o} </math>. Set <math>\sec(x) + \tan(x) = \alpha</math> and csc(x)+cot(x) = <math>\beta</math>. Then, notice that <math>\alpha + \beta = \frac{h+o}{a} + \frac{h+a}{o} = \frac{oh+ah+o^2 + a^2}{oa} = \frac{h(o+a+h)}{oa}</math> ( This is because of the Pythagorean Theorem, recall <math>o^2 +a^2 = h^2</math>). But then notice that <math>\alpha \cdot \beta = \frac{(o+h)(a+h)}{oa} = \frac{oa +oh +ha +h^2}{oa} = 1+ \frac{h(o+a+h)}{oa} = 1+ \alpha + \beta</math>. From the information provided in the question, we can substitute <math>\alpha</math> for <math>\frac{22}{7}</math>. Thus, <math>\frac{22 \beta}{7}= \beta + \frac{29}{7} \Longrightarrow 22 \beta = 7 \beta + 29 \Longrightarrow 15 \beta = 29 \Longrightarrow \beta = \frac{29}{15}</math>. Since, essentially we are asked to find the sum of the numerator and denominator of <math>\beta</math>, we have <math>29 + 15 = \boxed{044}</math>.
+~qwertysri987
+== Solution 8 ==
+Firstly, we write <math>\sec x+\tan x=a/b</math> where <math>a=22</math> and <math>b=7</math>. This will allow us to spot factorable expressions later. Now, since <math>\sec^2x-\tan^2x=1</math>, this gives us <cmath>\sec x-\tan x=\frac{b}{a}</cmath> Adding this to our original expressions gives us <cmath>2\sec x=\frac{a^2+b^2}{ab}</cmath> or <cmath>\cos x=\frac{2ab}{a^2+b^2}</cmath> Now since <math>\sin^2x+\cos^2x=1</math>, <math>\sin x=\sqrt{1-\cos^2x}</math> So we can write <cmath>\sin x=\sqrt{1-\frac{4a^2b^2}{(a^2+b^2)^2}}</cmath> Upon simplification, we get <cmath>\sin x=\frac{a^2-b^2}{a^2+b^2}</cmath> We are asked to find <math>1/\sin x+\cos x/\sin x</math> so we can write that as <cmath>\csc x+\cot x=\frac{1}{\sin x}+\frac{\cos x}{\sin x}</cmath> <cmath>\csc x+\cot x=\frac{a^2+b^2}{a^2-b^2}+\frac{2ab}{a^2+b^2}\frac{a^2+b^2}{a^2-b^2}</cmath> <cmath>\csc x+\cot x=\frac{a^2+b^2+2ab}{a^2-b^2}</cmath> <cmath>\csc x+\cot x=\frac{(a+b)^2}{(a-b)(a+b)}</cmath> <cmath>\csc x+\cot x=\frac{a+b}{a-b}</cmath> Now using the fact that <math>a=22</math> and <math>b=7</math> yields, <cmath>\csc x+\cot x=\frac{29}{15}=\frac{p}{q}</cmath> so <math>p+q=15+29=\boxed{44}</math>
+~Chessmaster20000
+== Solution 9 ==
+Rewriting <math>\sec{x}</math> and <math>\tan{x}</math> in terms of <math>\sin{x}</math> and <math>\cos{x}</math>, we know that <math>\frac{1+\sin{x}}{\cos{x}}=\frac{22}{7}.</math>
+Clearing fractions,
+<cmath>22\cos{x}=7+7\sin{x}.</cmath>
+Squaring to get an expression in terms of <math>\sin^2{x}</math> and <math>\cos^2{x}</math>,
+<cmath>484\cos^2{x}=49+49\sin^2{x}+98\sin{x}.</cmath>
+Substituting <math>\cos^2{x}=1-\sin^2{x},</math>
+<cmath>484(1-\sin^2{x})=49+49\sin^2{x}+98\sin{x}.</cmath>
+Expanding then collecting terms yields a quadratic in <math>\sin{x}:</math>
+<cmath>533\sin^2{x}+98\sin{x}-435=0.</cmath>
+To make calculations easier, let <math>y=\sin{x}.</math>
+<cmath>533y^2+98y-435=0.</cmath>
+Upon inspection, <math>y=-1</math> is a root. Dividing by <math>y+1</math>,
+<cmath>533y^2+98y-435=(533y-435)(y+1).</cmath>
+Substituting <math>y=\sin{x},</math> we see that <math>\sin{x}=-1</math> doesn't work, as <math>\cos{x}=0</math>, leaving <math>\tan{x}</math> undefined.
+We conclude that <math>\sin{x}=\frac{435}{533}.</math>
+Since <math>\sin^2{x}+\cos^2{x}=1,</math>
+<cmath>\cos{x}=\pm \sqrt{\frac{533^2-435^2}{533^2}}.</cmath>
+<cmath>=\pm \frac{308}{533}.</cmath>
+After checking via the given equation, we know that only the positive solution works.
+Therefore,
+<cmath>\csc{x}+\cot{x}=\frac{1}{\sin{x}}+\frac{\cos{x}}{\sin{x}}</cmath>
+<cmath>=\frac{533}{435}+\frac{308}{435}</cmath>
+<cmath>=\frac{29}{15}=\frac{m}{n}.</cmath>
-If we square <math>\sec x = \frac{22}{7} - \tan x</math>, we find that <math>\sec^2 x = \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x</math>, so <math>1 = \left(\frac{22}7\right)^2 - \frac{44}7 \tan x</math>. Solving shows that <math>\tan x = \frac{435}{308}</math>.
+Adding <math>m</math> and <math>n</math>, our answer is <math>\boxed{\textbf{044}}.</math>
-Call <math>y = \frac mn</math>. Rewrite the second equation in a similar fashion: <math>\displaystyle 1 = y^2 - 2y\cot x</math>. Substitute in <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> to get a [[quadratic equation|quadratic]]: <math>0 = y^2 - \frac{616}{435} - 1</math>. The quadratic is [[factor]]able (though somewhat ugly); <math>(15y - 29)(29y + 15) = 0</math>. It turns out that only the [[positive]] root will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = 044</math>.
+-Benedict T (countmath1)
 == See also ==
-{{AIME box|year=1991|num-b=8|num-a=10}}
+{{AIME box|year=1991|num-b=8|num-a=10|t=8322}}
 [[Category:Intermediate Trigonometry Problems]]
+{{MAA Notice}}

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