# 2007 iTest Problems/Problem 17

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## 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 $\tan{x}$.

$\text{(A) }\frac{37\sqrt{2}-18}{71}\qquad \text{(B) }\frac{35\sqrt{2}-6}{71}\qquad \text{(C) }\frac{35\sqrt{3}+12}{33}\qquad \text{(D) }\frac{37\sqrt{3}+24}{33}\qquad$

$\text{(E) }1\qquad \text{(F) }\frac{5}{7}\qquad \text{(G) }\frac{3}{7}\qquad \text{(H) }6\qquad$

$\text{(I) }\frac{1}{6}\qquad \text{(J) }\frac{1}{2}\qquad \text{(K) }\frac{6}{7}\qquad \text{(L) }\frac{4}{7}\qquad \text{(M) }\sqrt{3}\qquad$

$\text{(N) }\frac{\sqrt{3}}{3}\qquad \text{(O) }\frac{5}{6}\qquad \text{(P) }\frac{2}{3}\qquad \text{(Q) }\frac{1}{2007}\qquad$

## Solution

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

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

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:

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

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

Rearranging and solving, we get

$\tan{x}=\boxed{\textbf{(F) } \frac{5}{7}}$

## Solution 2

We will use complex numbers: $y$ represents the complex number $6+i$, and $x+y=\frac{\pi}{4}$ represents the fact $x\cdot y=1+i$, so dividing $1+i$ by $6+i$, we get $\frac{7+5i}{56}$, which means $\tan{x}=\boxed{\textbf{(F) } \frac{5}{7}}$