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

Revision as of 05:31, 30 July 2016

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

From the second equation, we get that $y=\arctan\frac{1}{6}$ . Plugging this into the first equation, we get:

$x+\arctan\frac{1}{6}=\frac{\pi}{4}$

Taking the tangent of both sides,

$\tan(x+\arctan\frac{1}{6})=\tan\frac{\pi}{4}=1$

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

$\tan{x}=\boxed{\frac{5}{7}}$

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 == 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>\tan{(x+/arctan{\frac{1}{6})}=\tan{\frac{\pi}{4}=1</math>. From the tangent addition formula, we then get:
+<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}+\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>.
-<math>\tan{x}=\box{\frac{5}{7}}</math>
+Rearranging and solving, we get
+<math>\tan{x}=\boxed{\frac{5}{7}}</math>