# 2008 iTest Problems/Problem 57

## Problem

Let a and b be the two possible values of $\tan\theta$ given that $\sin\theta + \cos\theta = \dfrac{193}{137}$. If $a+b=m/n$, where $m$ and $n$ are relatively prime positive integers, compute $m+n$.

## Solution

Let $x = \tfrac{193}{137}$. Multiply both sides by $\tfrac{\sqrt{2}}{2}$ to get $$\frac{\sqrt{2}}{2}\sin\theta + \frac{\sqrt{2}}{2}\cos\theta = \frac{\sqrt{2}}{2}x$$ $$\sin(\theta + 45^\circ) = \frac{\sqrt{2}}{2}x$$ $$\theta = \sin^{-1} (\frac{\sqrt{2}}{2}x) - 45^\circ \text{ or } 135 - \sin^{-1} (\frac{\sqrt{2}}{2}x)$$ Let $y = \tan(\sin^{-1} (\tfrac{\sqrt{2}}{2}x))$. Using the sum formula for tangent, the sum of the two possible values of $\tan \theta$ is $$\left(\frac{y - 1}{1 + y} \right) + \left(\frac{-1 - y}{1 - y} \right)$$ $$\left(\frac{y - 1}{1 + y} \right) + \left(\frac{y+1}{y-1} \right)$$ $$\left(\frac{y^2 - 2y + 1}{y^2 - 1} \right) + \left(\frac{y^2 + 2y + 1}{y^2 - 1} \right)$$ $$\frac{2y^2 + 2}{y^2 - 1}$$ If a triangle has opposite side $x\sqrt{2}$ and hypotenuse $2$, by the Pythagorean Theorem, the adjacent side's length is $\sqrt{4-2x^2}$. That means if $\sin(\alpha) = \tfrac{\sqrt{2}}{2}x$, then $\tan(\alpha) = \sqrt{\tfrac{2x^2}{4-2x^2}}$. That means $y = \tan(\sin^{-1} (\tfrac{\sqrt{2}}{2}x)) = \sqrt{\tfrac{2x^2}{4-2x^2}}$. Therefore, the sum of the two possible values of $\tan \theta$ is $$\frac{2 \cdot \frac{2x^2}{4-2x^2} + 2}{\frac{2x^2}{4-2x^2} - 1}$$ $$\frac{ \frac{4x^2}{4-2x^2} + \frac{8-4x^2}{4-2x^2} }{\frac{2x^2}{4-2x^2} - \frac{4 - 2x^2}{4-2x^2} }$$ $$\frac{8}{4x^2-4}$$ $$\frac{2}{x^2 - 1}$$ $$\frac{2}{(x+1)(x-1)}$$ $$\frac{2}{\frac{330}{137} \cdot \frac{56}{137}}$$ $$\frac{137 \cdot 137}{330 \cdot 28}$$ $$\frac{18769}{9240}$$ Thus, $m+n = \boxed{28009}$.