2007 iTest Problems/Problem 17: Difference between revisions

Revision as of 05:33, 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:

$\dfrac{\tan{x}+\frac{1}{6}}{1-\frac{1}{6}\tan{x}}=1$

$\tan{x}+\frac{1}{6}=1-\frac{1}{6}\tan{x}$ .

Rearranging and solving, we get

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

@@ Line 14: / Line 14: @@
 From the tangent addition formula, we then get:
-<math>\frac{\tan{x}+\frac{1}{6}}{1-\frac{1}{6}\tan{x}}=1</math>
+<math>\dfrac{\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>.