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

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__TOC__
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== Problem ==
 
== 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>
 
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>
  
−
__TOC__
+
== Solution 1 ==
−
== Solution ==
 
−
=== Solution 1 ===
 
 
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>.  
 
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>.  
  
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Let <math>y = \frac mn</math>. Then squaring,  
 
Let <math>y = \frac mn</math>. Then squaring,  
  
−
<cmath>\csc^2 x = (y - \cot^2 x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.</cmath>  
+
<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>.
 
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>.
  
−
=== Solution 2 ===
+
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
 
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
  
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so <math>m + n = 044</math>.
 
so <math>m + n = 044</math>.
  
−
=== Solution 3 ===
+
== Solution 3 (least computation)==
−
(least computation) By the given,
+
By the given,
 
<math>\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}</math> and
 
<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>.
 
<math>\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k</math>.
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<cmath>1 = \frac {22}{7}k - \frac {22}{7} - k.</cmath>
 
<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>  
+
Solving yields <math>k = \frac {29}{15}</math>, and <math>m+n = 044</math>
  
−
=== Solution 4 ===
+
== Solution 4 ==
−
Make the substitution <math>u = \tan \frac x2</math> (a substitution commonly used in calculus). <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>. <math>\sec x + \tan x = \frac{1 + \sin x}{\cos x}.</math> Now note the following:
+
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}\\
 
<cmath>\begin{align*}\sin x &= \frac{2u}{1+u^2}\\
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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>.
 
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 ===
+
== 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>
 
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>=\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>\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>.
 
<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>
 +
 +
Adding <math>m</math> and <math>n</math>, our answer is <math>\boxed{\textbf{044}}.</math>
 +
 +
-Benedict T (countmath1)
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Trigonometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]
 +
{{MAA Notice}}

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)

See also

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