Working with Tangents

Write a solution.

Let $a = \dfrac{\sin(x)}{\cos(y)}$ and $b = \dfrac{\sin(y)}{\cos(x)}.$ Then we can write the system of equations as

$a + b = 1, \dfrac{1}{a} + \dfrac{1}{b} = 6 \Longrightarrow \dfrac{1}{a} + \dfrac{1}{1 - a} = 6 \Longrightarrow 1 = 6a(1 - a)$

$\Longrightarrow 6a^{2} - 6a + 1 = 0 \Longrightarrow a = \dfrac{6 \pm \sqrt{36 - 24}}{12} = \dfrac{3 \pm \sqrt{3}}{6}.$

Then $b = 1 - a = \dfrac{3 \mp \sqrt{3}}{6}.$ Because of this symmetry, we can, without loss of generality, let $a = \dfrac{3 + \sqrt{3}}{6}$ and $b = \dfrac{3 - \sqrt{3}}{6}.$

This gives us $ab = \tan(x)\tan(y) = \dfrac{6}{36} = \dfrac{1}{6}$ , and so

$\dfrac{\tan(x)}{\tan(y)} + \dfrac{\tan(y)}{\tan(x)} = \dfrac{\tan^{2}(x) + \tan^{2}(y)}{\tan(x)\tan(y)} = 6*(\sec^{2}(x) + \sec^{2}(y) - 2),$ (A).

Now $a = \dfrac{\sin(x)}{\cos(y)} = \dfrac{3 + \sqrt{3}}{6} \Longrightarrow \sin(x) = \dfrac{3 + \sqrt{3}}{6}\cos(y),$ and

$b = \dfrac{\sin(y)}{\cos(x)} = \dfrac{3 - \sqrt{3}}{6} \Longrightarrow \cos(x) = \dfrac{6}{3 - \sqrt{3}}\sin(y) = (3 + \sqrt{3})\sin(y).$

Thus $\sin^{2}(x) + \cos^{2}(x) = \dfrac{2 + \sqrt{3}}{6}\cos^{2}(y) + (12 + 6\sqrt{3})\sin^{2}(y)$

$\Longrightarrow 1 = \dfrac{2 + \sqrt{3}}{6}\cos^{2}(y) + (12 + 6\sqrt{3})(1 - \cos^{2}(y))$

$\Longrightarrow \cos^{2}(y)\left(12 + 6\sqrt{3} - \dfrac{2 + \sqrt{3}}{6}\right) = 11 + 6\sqrt{3}$

$\Longrightarrow \sec^{2}(y) = \dfrac{35(2 + \sqrt{3})}{6(11 + 6\sqrt{3})} = \dfrac{35(4 - \sqrt{3})}{78}.$

Now $a^{2} = \sin^{2}(x)\sec^{2}(y) = \dfrac{2 + \sqrt{3}}{6},$ so

$\sin^{2}(x) = \dfrac{2 + \sqrt{3}}{6} * \dfrac{78}{35(4 - \sqrt{3})} = \dfrac{11 + 6\sqrt{3}}{35},$ and thus

$\cos^{2}(x) = 1 - \sin^{2}(x) = \dfrac{24 - 6\sqrt{3}}{35}$

$\Longrightarrow \sec^{2}(x) = \dfrac{35}{6(4 - \sqrt{3})} = \dfrac{35(4 + \sqrt{3})}{78}.$

Thus equation (A) becomes

$6*\left(\dfrac{35(4 - \sqrt{3})}{78} + \dfrac{35(4 + \sqrt{3})}{78} - 2\right) = 6*\left(\dfrac{280}{78} - 2\right) = \dfrac{124}{13},$

and so $p + q = 124 + 13 = \boxed{137}.$

Let $a=\frac{\cos x}{\sin y}$ and $b=\frac{\cos y}{\sin x}$ thus from $\frac{\sin x}{\cos y} +\frac{\sin x}{\cos y} =1$ and $\frac{\cos x}{\sin y} +\frac{\cos y}{\sin x}=6$ . $a+b=6, \frac{1}{a}+\frac{1}{b}=1$ . This leads to $\frac{1}{b}+\frac{1}{6-b}=1$ . thus $6=6b-b^2$ and $b=3 \pm \sqrt3$ . Let $b=3+\sqrt3$ and $a=3-\sqrt3$ .

$\frac{\cos x}{\sin y}= 3-\sqrt3$ and $\frac{\cos y}{\sin x}=3+\sqrt3$

so $\cos x=(3-\sqrt3)\sin y$ and $\cos y = (3+\sqrt3)\sin x$

$\cos^2 x+ \sin^2 x=1, \cos^2 y+ \sin^2 y=1$ . substitute x's for y's and y's for x's to get two equations.

$( (3-\sqrt3)\sin y)^2+ (\frac{\cos y}{3+\sqrt3})^2=1$ and $( (3+\sqrt3)\sin x)^2 + (\frac{\cos x}{3-\sqrt3})^2 =1$

Note $\frac{1}{3-\sqrt3}= \frac{1}{6}(3+\sqrt3)$ and $\frac{1}{3+\sqrt3}=\frac{1}{6}(3-\sqrt3)$

We then simplify these equations and solve for $\sin x$ and $\sin y$

$(3-\sqrt3)^2(\sin^2 y +\frac{1}{36}\cos^2 y)=1$

$(3+\sqrt3)^2(\sin^2x + \frac{1}{36}\cos^2 x)=1$

$\Longrightarrow$

$\sin^2 y +\frac{1}{36}(1-\sin^2 y)=\frac{1}{(3-\sqrt3)^2}$

$\sin^2 x +\frac{1}{36}(1-\sin^2 x)=\frac{1}{(3+\sqrt3)^2}$

$\Longrightarrow$

$\frac{35}{36}\sin^2 y +\frac{1}{36}=\frac{1}{6}(2-\sqrt3)$

$\frac{35}{36}\sin^2 x +\frac{1}{36}=\frac{1}{6}(2+\sqrt3)$

$\Longrightarrow$

$\frac{35}{36}\sin^2 y=\frac{11+6\sqrt3}{36}$

$\frac{35}{36}\sin^2 x =\frac{11-6\sqrt3}{36}$

$\Longrightarrow$

$\sin y=\sqrt\frac{11+6\sqrt3}{35}$

$\sin x =\sqrt\frac{11-6\sqrt3}{35}$

Solve for $\cos y$ and $\cos x$

$\cos x=(3-\sqrt3)\sin y$ and $\cos y = (3+\sqrt3)\sin x$

$\cos y = ( 3+\sqrt3)\sqrt\frac{11-6\sqrt3}{35}$

$\cos x = ( 3-\sqrt3)\sqrt\frac{11+6\sqrt3}{35}$

These reduce to

$\cos y = \sqrt\frac{6(4-\sqrt3)}{35}$

$\cos x = \sqrt\frac{6(4+\sqrt3)}{35}$ .

$\frac{\tan x}{\tan y} + \frac{\tan y}{\tan x}=\frac{p}{q}$

Let $\frac{\tan x}{\tan y}=z$ solve for $z$

$z= \frac{\tan x}{\tan y} = \frac{\sin x\cos y}{\cos x\sin y}=\frac{\sqrt\frac{11-6\sqrt3}{35} \sqrt\frac{6(4-\sqrt3)}{35}}{ \sqrt\frac{11+6\sqrt3}{35}\sqrt\frac{6(4+\sqrt3)}{35}}=\sqrt{\frac{11-6\sqrt3}{11+6\sqrt3}\cdot \frac{4-\sqrt3}{4+\sqrt3}}=\sqrt\frac{7519-4340\sqrt3}{169}$ $z=\frac{62-35\sqrt3}{13}$

$z+\frac{1}{z}=\frac{p}{q}$

$\frac{p}{q}=\frac{62-35\sqrt3}{13}+\frac{13}{62-35\sqrt3}=\frac{(62-35\sqrt3)^2+13^2}{13(62-35\sqrt3)}=\frac{7688-4340\sqrt3}{13(62-35\sqrt3)} =\frac{124(62-35\sqrt3)}{13(62-35\sqrt3)}=\frac{124}{13}$

$p+q=124+13=\boxed{137}$