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

Revision as of 13:38, 6 May 2007

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

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

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

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

Using the tangent addition formula:

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

@@ Line 13: / Line 13: @@
 <math>\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = 150</math>
 == See also ==
-* [[1986 AIME Problems]]
 {{AIME box|year=1986|num-b=2|num-a=4}}
+* [[AIME Problems and Solutions]]
+* [[American Invitational Mathematics Examination]]
+* [[Mathematics competition resources]]

1986 AIME (Problems • Answer Key • Resources)
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All AIME Problems and Solutions