Difference between revisions of "1986 AIME Problems/Problem 3"
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If <math>\tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>? | If <math>\tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>? | ||
− | == Solution == | + | == Solution 1 == |
Since <math>\cot</math> is the reciprocal function of <math>\tan</math>: | Since <math>\cot</math> is the reciprocal function of <math>\tan</math>: | ||
Line 11: | Line 11: | ||
Using the tangent addition formula: | Using the tangent addition formula: | ||
− | <math>\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = 150</math> | + | <math>\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = \boxed{150}</math>. |
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+ | == Solution 2 == | ||
+ | |||
+ | Using the formula for tangent of a sum, <math>\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y} = \frac{25}{1-\tan x \tan y}</math>. We only need to find <math>\tan x \tan y</math>. | ||
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+ | We know that <math>25 = \tan x + \tan y = \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}</math>. Cross multiplying, we have <math>\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} = \frac{\sin(x+y)}{\cos x \cos y} = 25</math>. | ||
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+ | Similarly, we have <math>30 = \cot x + \cot y = \frac{\cos x}{\sin x} + \frac{\cos y}{\sin y} = \frac{\cos x \sin y + \sin x \cos y}{\sin x \sin y} = \frac{\sin(x+y)}{\sin x \sin y}</math>. | ||
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+ | Dividing: | ||
+ | |||
+ | <math>\frac{25}{30} = \frac{\frac{\sin(x+y)}{\cos x \cos y}}{\frac{\sin(x+y)}{\sin x \sin y}} = \frac{\sin x \sin y}{\cos x \cos y} = \tan x \tan y = \frac{5}{6}</math>. Plugging in to the earlier formula, we have <math>\tan(x+y) = \frac{25}{1-\frac{5}{6}} = \frac{25}{\frac{1}{6}} = \boxed{150}</math>. | ||
== See also == | == See also == |
Latest revision as of 13:43, 15 April 2017
Contents
Problem
If and , what is ?
Solution 1
Since is the reciprocal function of :
Thus,
Using the tangent addition formula:
.
Solution 2
Using the formula for tangent of a sum, . We only need to find .
We know that . Cross multiplying, we have .
Similarly, we have .
Dividing:
. Plugging in to the earlier formula, we have .
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
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