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

Revision as of 17:19, 13 February 2008

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 $\tan$ :

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

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

Using the tangent addition formula:

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

@@ Line 5: / Line 5: @@
 Since <math>\cot</math> is the reciprocal function of <math>\tan</math>:
-<math>\displaystyle \cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan x} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30</math>
+<math>\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30</math>
 Thus, <math>\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}</math>
@@ Line 12: / Line 12: @@
 <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 ==
 {{AIME box|year=1986|num-b=2|num-a=4}}

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