Difference between revisions of "1986 AIME Problems/Problem 3"

Revision as of 18:04, 4 July 2013

If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$ ?

Since $\cot$ is the reciprocal function of $\tan$ :

$\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30$

Thus, $\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}$

Using the tangent addition formula:

$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = 150$

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 == Problem ==
-If <math>\displaystyle \tan x+\tan y=25</math> and <math>\displaystyle \cot x + \cot y=30</math>, what is <math>\displaystyle \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 ==
@@ Line 18: / Line 18: @@
 * [[American Invitational Mathematics Examination]]
 * [[Mathematics competition resources]]
+{{MAA Notice}}

1986 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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