Difference between revisions of "2007 iTest Problems/Problem 17"

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(Solution)
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== Solution ==
 
== Solution ==
From the second equation, we get that <math>y=arctan{\frac{1}{6}}</math>. Plugging this into the first equation, we get:
+
From the second equation, we get that <math>y=arctan\frac{1}{6}</math>. Plugging this into the first equation, we get:
 
<math>x+arctan{\frac{1}{6}=\frac{\pi}{4}</math>. Taking the tangent of both sides,
 
<math>x+arctan{\frac{1}{6}=\frac{\pi}{4}</math>. Taking the tangent of both sides,
 
<math>\tan{(x+/arctan{\frac{1}{6})}=\tan{\frac{\pi}{4}=1</math>. From the tangent addition formula, we then get:
 
<math>\tan{(x+/arctan{\frac{1}{6})}=\tan{\frac{\pi}{4}=1</math>. From the tangent addition formula, we then get:
 
<math>\tan{x}+\frac{1}{6}/1-\frac{1}{6}\tan{x}=1</math>
 
<math>\tan{x}+\frac{1}{6}/1-\frac{1}{6}\tan{x}=1</math>
 +
 
<math>\tan{x}+\frac{1}{6}=1-\frac{1}{6}\tan{x}</math>. Rearranging and solving, we get:
 
<math>\tan{x}+\frac{1}{6}=1-\frac{1}{6}\tan{x}</math>. Rearranging and solving, we get:
 
<math>\tan{x}=\box{\frac{5}{7}}</math>
 
<math>\tan{x}=\box{\frac{5}{7}}</math>

Revision as of 06:25, 30 July 2016

Problem

If $x$ and $y$ are acute angles such that $x+y=\frac{\pi}{4}$ and $\tan{y}=\frac{1}{6}$, find the value of $\tan{x}$.

Solution

From the second equation, we get that $y=arctan\frac{1}{6}$. Plugging this into the first equation, we get: $x+arctan{\frac{1}{6}=\frac{\pi}{4}$ (Error compiling LaTeX. Unknown error_msg). Taking the tangent of both sides, $\tan{(x+/arctan{\frac{1}{6})}=\tan{\frac{\pi}{4}=1$ (Error compiling LaTeX. Unknown error_msg). From the tangent addition formula, we then get: $\tan{x}+\frac{1}{6}/1-\frac{1}{6}\tan{x}=1$

$\tan{x}+\frac{1}{6}=1-\frac{1}{6}\tan{x}$. Rearranging and solving, we get: $\tan{x}=\box{\frac{5}{7}}$ (Error compiling LaTeX. Unknown error_msg)