Sine Sine Cosine Cosine, I got it!

It is given that:

$\begin{cases} \dfrac {\sin{x}}{\sin{y}} = \dfrac{1}{2} & \Rightarrow \sin{y} = 2 \sin{x} & ...(1) \\ \dfrac {\cos{x}}{\cos{y}} = \dfrac{3}{2} & \Rightarrow \cos{y} = \dfrac {2}{3} \cos{x} & ...(2) \end{cases}$

$\begin{cases} Eq.1^2: & \Rightarrow \sin^2{y} = 4 \sin^2 {x} & ...(1a) \\ Eq.2^2: & \Rightarrow \cos^2{y} = \dfrac {4}{9} \cos^2{x} & ...(2a) \end{cases}$

$Eq.1a+Eq.2a: \quad \Rightarrow 1 = 4 \sin^2 {x} + \dfrac {4}{9} \cos^2{x} \\ \times \dfrac {1}{\cos^2{x}}: \quad \Rightarrow 4 \tan^2{x} + \dfrac {4} {9} = \sec^2{x} = 1 + \tan^2{x} \\ \Rightarrow 3\tan^2{x} = \dfrac {5}{9} \quad \Rightarrow \tan{x} = \dfrac {\sqrt{15}}{9}$

$\dfrac {Eq.1}{Eq.2}: \quad \Rightarrow \tan{y} = 3\tan{x}$

$\tan{(x+y)} = \dfrac {\tan{x}+\tan{y}}{1-\tan{x}\tan{y}} = \dfrac {4\tan{x}} {1-3\tan^2{x}} = \dfrac {\frac {4\sqrt{15}}{9}}{1-\frac{45}{81}} = \dfrac{36\sqrt{15}}{36} = \sqrt{15}$

Therefore, $\tan^2{(x+y)} = \boxed{15}$

Summing the equations we get

$\displaystyle{\dfrac{\sin x}{\sin y} + \dfrac{\cos x}{\cos y} = \dfrac{1}{2} + \dfrac{3}{2} \quad \Rightarrow \quad \dfrac{\sin x \cos y + \sin y \cos x}{\sin y \cos y}= 2}$ that is $\displaystyle{\sin x \cos y + \sin y \cos x = 2\sin y \cos y}$ hence we land on the $\color{#3D99F6}{very\ nice}$ equation $\boxed{\displaystyle{\sin(x+y) = \sin(2y)}}$ Since $0^\circ < x,y < 90^\circ$ , both $x+y$ and $2y$ are in the $(0^\circ, 180^\circ)$ range and the above equality then implies $x+y = 2y \qquad \textrm{OR} \qquad x+y = 180^\circ - 2y$ if $x+y=2y$ we'd get $x = y$ and ${\sin x \over \sin y} = 1$ against the first equality in our assumptions. So necessarily we have $x+y = 180^\circ - 2y \Rightarrow \displaystyle{\boxed{x = 180^\circ - 3y\ }}$ As consequence we have two nice facts. The first is

$\displaystyle{ \dfrac{\sin (180^\circ - 3y)}{\sin y} = \dfrac{1}{2} \Rightarrow \dfrac{\sin (3y)}{\sin y} = \dfrac{\sin (2y)\cos y + \cos(2y)\sin y}{\sin y} = \dfrac{1}{2}}$ and so $\displaystyle{ 2\cos^2y + 2\cos^2y - 1 = 4\cos^2y -1 = \dfrac{1}{2} \Rightarrow \cos y = \sqrt{\dfrac{3}{8}} }$ that also means $\sin y = \sqrt{\dfrac{5}{8}}$ (in derivinq these values we used the fact that $0 < y < 90^\circ$ so both $\sin y$ and $\cos y$ are positive).

The second thing is

$\tan(x+y) = \tan(180^\circ - 2y) = -\tan(2y) =$ $\displaystyle{= \dfrac{2 \tan y}{\tan^2 x - 1} = \dfrac{2 \sqrt{\dfrac{5}{3}}} {\dfrac{5}{3} - 1} = \dfrac{ 2 \sqrt{\dfrac{5}{3}}} {\dfrac{2}{3} } =3 \sqrt{\dfrac{5}{3}} = \sqrt{15} }$

hence

$\boxed{\tan^2(x+y) = 15}$

See: componendo and dividendo

To save time and space, let's use s,c and t to stand for sin, cos and tan.

The first equation gives

$sx=\dfrac{sy}{2}\dots(1)$

and the second equation gives

$cx=\dfrac{3cy}{2}\dots(2)$

Squaring these equations and adding gives

$\dfrac{s^2y}{4}+\dfrac{9c^2y}{4}=1$

From this it follows easily that

$s^2y=\dfrac{5}{8}$

and so

$c^2y=\dfrac{3}{8}$

Now using equations (1) and (2) it is easy to find

$s^2x=\dfrac{5}{32}$

$c^2x=\dfrac{27}{32}$

Using these last four equations we can find

$t^2{y}=\dfrac{5}{3}$

$t^2{x}=\dfrac{5}{27}$

Using these last two equations in the standard formula for $\tan{(x+y)}$ gives

$t^2(x+y)=\left( \dfrac{\sqrt{\frac{5}{3}}+\sqrt{\frac{5}{27}}}{1-\sqrt{\frac{5}{3}}\sqrt{\frac{5}{27}}}\right)^2$

After a couple of lines of careful arithmetic this simplifies to $\boxed{15}$