{
    "problem": "If\n\\[\\frac{\\sin x}{\\cos y} + \\frac{\\sin y}{\\cos x} = 1 \\quad \\text{and} \\quad \\frac{\\cos x}{\\sin y} + \\frac{\\cos y}{\\sin x} = 6,\\]then find $\\frac{\\tan x}{\\tan y} + \\frac{\\tan y}{\\tan x}.$",
    "level": "Level 5",
    "type": "Precalculus",
    "solution": "From the first equation,\n\\[\\frac{\\sin x \\cos x + \\sin y \\cos y}{\\cos x \\cos y} = 1.\\]From the second equation,\n\\[\\frac{\\cos x \\sin x + \\cos y \\sin y}{\\sin x \\sin y} = 6.\\]Dividing these equations, we get\n\\[\\tan x \\tan y = \\frac{1}{6}.\\]Multiplying the two given equations, we get\n\\[\\frac{\\sin x \\cos x}{\\sin y \\cos y} + 1 + 1 + \\frac{\\sin y \\cos y}{\\sin x \\cos x} = 6,\\]so\n\\[\\frac{\\sin x \\cos x}{\\sin y \\cos y} + \\frac{\\sin y \\cos y}{\\sin x \\cos x} = 4.\\]Note that\n\\begin{align*}\n\\sin x \\cos x &= \\frac{\\sin x \\cos x}{\\sin^2 x + \\cos^2 x} \\\\\n&= \\frac{\\frac{\\sin x}{\\cos x}}{\\frac{\\sin^2 x}{\\cos^2 x} + 1} \\\\\n&= \\frac{\\tan x}{\\tan^2 x + 1}.\n\\end{align*}Similarly, $\\sin y \\cos y = \\frac{\\tan y}{\\tan^2 y + 1},$ so\n\\[\\frac{\\tan x (\\tan^2 y + 1)}{\\tan y (\\tan^2 x + 1)} + \\frac{\\tan y (\\tan^2 x + 1)}{\\tan x (\\tan^2 y + 1)} = 4.\\]Then\n\\[\\frac{\\tan x \\tan^2 y + \\tan x}{\\tan y \\tan^2 x + \\tan y} + \\frac{\\tan y \\tan^2 x + \\tan y}{\\tan x \\tan^2 y + \\tan x} = 4.\\]Since $\\tan x \\tan y = \\frac{1}{6},$\n\\[\\frac{\\frac{1}{6} \\tan y + \\tan x}{\\frac{1}{6} \\tan x + \\tan y} + \\frac{\\frac{1}{6} \\tan x + \\tan y}{\\frac{1}{6} \\tan y + \\tan x} = 4.\\]Thus,\n\\[\\frac{\\tan y + 6 \\tan x}{\\tan x + 6 \\tan y} + \\frac{\\tan x + 6 \\tan y}{\\tan y + 6 \\tan x} = 4.\\]Then\n\\[(\\tan y + 6 \\tan x)^2 + (\\tan x + 6 \\tan y)^2 = 4 (\\tan x + 6 \\tan y)(\\tan y + 6 \\tan x),\\]or\n\\begin{align*}\n&\\tan^2 y + 12 \\tan x \\tan y + 36 \\tan^2 x + \\tan^2 x + 12 \\tan x \\tan y + 36 \\tan^2 y \\\\\n&= 4 \\tan x \\tan y + 24 \\tan^2 x + 24 \\tan^2 y + 144 \\tan x \\tan y.\n\\end{align*}This reduces to\n\\[13 \\tan^2 x + 13 \\tan^2 y = 124 \\tan x \\tan y = \\frac{124}{6},\\]so $\\tan^2 x + \\tan^2 y = \\frac{62}{39}.$\n\nFinally,\n\\[\\frac{\\tan x}{\\tan y} + \\frac{\\tan y}{\\tan x} = \\frac{\\tan^2 x + \\tan^2 y}{\\tan x \\tan y} = \\frac{\\frac{62}{39}}{\\frac{1}{6}} = \\boxed{\\frac{124}{13}}.\\]"
}