1 Trillion Followers Problem

I'm going straight for the generalization.

$\begin{aligned} \tan x + \tan y &=&\frac{\sin x }{\cos x} + \frac{ \sin y}{\cos y} \\ & = & \frac { \sin x \cos y + \cos x \sin y}{\cos x \cos y} \\ & = & \frac{2 \sin(x+y)}{2\cos x \cos y} \\ & = & \color{#3D99F6}{\frac{2 \sin(x+y)}{\cos(x+y) + \cos(x-y)} }\\ \end{aligned}$

From the first equation:

$\begin{aligned} \sin(2x) + \sin(2y) & = & A \\ 2 \sin(x+ y) \cos(x-y) & = & A \\ \color{#20A900}{ \sin(x+ y)} & = & \color{#20A900}{\frac A2 \sec(x-y)} \\ \end{aligned}$

From the second equation:

$\begin{aligned} \cos(2x) + \cos(2y) & = & B \\ 2 \cos(x+ y) \cos(x-y) & = & B \\ \color{#69047E}{ \cos(x+ y)} & = & \color{#69047E}{\frac B2 \sec(x-y)} \\ \end{aligned}$

$\color{#20A900}{GREEN} ^2 + \color{#69047E}{PURPLE} ^2$ :

$\begin{aligned} \sin^2(x+ y) + \cos^2(x+y) &=& \left ( \frac {A^2}{4} + \frac {B^2}{4} \right) \sec^2 (x-y) \\ 1 &=& \frac 14 (A^2+B^2) \sec^2 (x-y) \\ \cos(x-y) &=& \pm \frac{\sqrt{A^2+B^2}}{2} \\ \end{aligned}$

Solve for $\color{#20A900}{GREEN}$ and $\color{#69047E}{PURPLE}$ gives

$\sin(x+y) = \pm \frac A{\sqrt{A^2+B^2}}$ and $\cos(x+y) = \pm \frac B{\sqrt{A^2+B^2}}$

When taking their ratios as in $\color{#3D99F6}{BLUE}$ , the signs "match up" and cancel out.

$\large \tan x + \tan y = \frac{2 \cdot \frac A{\sqrt{A^2+B^2}} } { \frac B{\sqrt{A^2+B^2}} + \frac{\sqrt{A^2+B^2}}2 } = \frac{4A}{2B + A^2+B^2} \\$

In this case, $A = \frac 13, B = \frac57 \Rightarrow \tan x + \tan y = \frac{147}{226}$ .

We can interpret the given equations as the addition of unit vectors, $(\cos{2x},\sin{2x})+(\cos{2y},\sin{2y})=\left(\frac{5}{7},\frac{1}{3}\right)$ . See the attached figure, where $r=\sqrt{A^2+B^2}$

We can find the points $(\cos{2x},\sin{2x})$ and $(\cos{2y},\sin{2y})$ as the two points of intersection of the unit circles centered at the origin and at the point $(A,B)$ . These points turn out to be $\frac{1}{2}(A,B)\pm\frac{\sqrt{1-r^2/4}}{r}(B,-A)$ ; this result should make good sense in terms of our figure. Now $\tan{x}+\tan{y}=\frac{1-\cos{2x}}{\sin{2x}}+\frac{1-\cos{2y}}{\sin{2y}}$ ...

Here is an ALTERNATIVE SOLUTION, more in line with the two others, but from a vectorial point of view (or just think in terms of polar coordinates!), supported by the attached figure. Let $r=\sqrt{A^2+B^2}$ , and note that the polar angle of the point $(A,B)$ is $x+y$ (we may have to change the angle $x$ by $\pi$ , which will not affect the problem). Now $\cos(x+y)=\frac{B}{r},\quad\sin(x+y)=\frac{A}{r},\quad\cos(y-x)=\frac{r}{2},$ $\tan{x}+\tan{y}=\frac{2\sin(x+y)}{\cos(x+y)+\cos(y-x)}=\frac{2A/r}{B/r+r/2}$ $=\frac{4A}{2B+r^2}=\frac{4A}{2B+A^2+B^2}=\boxed{\frac{147}{226}}$

$\sin {2x} + \sin{2y}=A$ $2\sin{\left( x+y \right) }\sin{\left( x-y \right)=A \quad \dots 1}$ $\cos {2x} + \cos{2y}=B$ $2\cos{\left( x+y \right) }\cos{\left( x-y \right) }=B \quad \dots 2$ Dividing equation $1$ by equation $2$ . $\tan{\left( x+y \right)=\frac{A}{B}} \dots 3$ $\tan{\left( x+y \right)}=\frac{\tan{x}+\tan{y}}{1-\tan{x} \tan{y}}$ $\tan{x}+\tan{y}=\tan{\left( x+y \right)}\left(1-\tan{x} \tan{y}\right) \quad \dots 4$ $\tan{x} \tan{y}=\frac{\sin{x}\sin{y}}{\cos{x}\cos{y}}$ $=\frac{2\sin{x}\sin{y}}{2\cos{x}\cos{y}}$ $=\frac{\cos{\left( x-y \right) }-\cos{\left( x+y \right) }}{\cos{\left( x+y \right) }-\cos{\left( x-y \right) }} \quad \dots 5$ Squaring and adding $1$ and $2$ , $4\cos ^{ 2 }{ \left( x-y \right) ={ A }^{ 2 }+{ B }^{ 2 } }$ $\cos { \left( x-y \right) } =\quad \frac { \sqrt { { A }^{ 2 }+{ B }^{ 2 } } }{ 2 } \quad \dots 6$ Using equation $2$ we have, $\cos { \left( x+y \right) } =\quad \frac { B }{ \sqrt { { A }^{ 2 }+{ B }^{ 2 } } } \quad \dots 7$ Using equation $6$ and $7$ in equation $5$ we get, $\tan { x } \tan { y } =\frac { { A }^{ 2 }+{ B }^{ 2 }-2B }{ { A }^{ 2 }+{ B }^{ 2 }+2B } \quad \dots 8$ Using equation $3$ and $8$ in $4$ , $\tan{x}+\tan{y}=\frac { 4A }{ { A }^{ 2 }+{ B }^{ 2 }+2B }$

$\large \boxed{\frac{147}{226}}$