Can a ratio of sinusoidal functions with the same frequency always be written as a tangent function? In general, two sinusoidal functions $y_1$ and $y_2$ with angular frequency $\omega$ can be written as

*

*$y_1 = A\sin(\omega x) + B \cos(\omega x)$,


*$y_2 = C\sin(\omega x) + D \cos(\omega x)$.
Where $A, B, C, D$ are real numbers and it is not the case that $A=B=0$ or $C=D=0$.
Can the ratio $\frac{y_1}{y_2}$ always be written in the form
$$ \frac{y_1}{y_2} = E \tan(\omega x + h) + k$$
for some constants $E, h$, and $k$?
Edit:
For example
$$\frac{12 \sin (x) + 3 \cos(x)}{  4 \sin(x) + 2 \cos(x)}$$
has a graph like this:

Which looks like the tangent function transformed by scaling and shifting.
 A: First let's dispense with having to worry about $\omega$ all the time.
Let $t = \omega x.$ Then
\begin{align}
y_1 &= A\sin(t) + B \cos(t),\\
y_2 &= C\sin(t) + D \cos(t).
\end{align}
Let $R$ and $\alpha$ be such that
$$ C\sin(t) + D \cos(t) = R \cos(t + \alpha). $$
Then
$$
R\sin(t + \alpha) = -\frac{\mathrm d}{\mathrm dt} R\cos(t + \alpha)
= D \sin(t) - C \cos(t).
$$
Now we want $u$ and $v$ such that
$u \langle D, -C \rangle + v \langle C, D\rangle = \langle A, B\rangle.$
Since $C^2 + D^2 \neq 0$ we can solve this as
$$
u = \frac{AD - BC}{C^2 + D^2}, \quad v = \frac{AC + BD}{C^2 + D^2}.
$$
Then
\begin{align}
\frac{y_1}{y_2}
&= \frac{u(D \sin(t) - C \cos(t)) + v(C\sin(t) + D \cos(t))}{C\sin(t) + D \cos(t)} \\
&= \frac{uR\sin(t + \alpha) + vR \cos(t + \alpha)}{R \cos(t + \alpha)} \\
&= u\tan(t + \alpha) + v \\
&= u\tan(\omega x + \alpha) + v.
\end{align}
A: Yes: for $C,D$ not both zero, $$\frac{A\sin \omega x+B\cos \omega x}{C\sin \omega x+D\cos \omega x}=\frac{AD-BC}{C^2+D^2}\;\tan\left(\omega x-\arctan\frac CD\right)+\frac{AC+BD}{C^2+D^2}.$$ This is true even when $D=0,$ if we let $\displaystyle\arctan\frac CD=\begin{cases}\displaystyle&\frac\pi2&\text{if }C>0;
\\ \displaystyle-&\frac\pi2 &\text{if }C<0.\end{cases}$

*

*The above
expressions
for the parameters $E,h,k$ were obtained by equating
$$\frac{A\sin \omega x+B\cos \omega x}{C\sin \omega x+D\cos \omega x}$$ and the given tangent function $$E \tan(\omega x + h) + k,$$ expanding the
latter into sines and cosines, observing from the denominators that
$$\frac CD=-\tan h,$$ and setting $$h=-\arctan\frac CD.$$
Now, $h$ is in either quadrant $4$ or $1$ depending on the relative
signs of $C$ & $D,$ so $$\sin h=\frac{\mp C}{C^2+D^2},\\\cos
   h=\frac{D}{C^2+D^2}.$$
Substituting these two equalities into the equation where we left off
gives $$\sqrt{C^2+D^2}(E\cos h-k\sin h)=A,\\\sqrt{C^2+D^2}(E\sin
   h+k\cos h)=B.$$ Solving this system finally gives the above
expressions for $E$ and $k.$

*Here's a rigorous proof: $$\frac{AD-BC}{C^2+D^2}\;\tan\left(\omega
   x-\arctan\frac CD\right)+\frac{AC+BD}{C^2+D^2}\\
   =\frac{AD-BC}{C^2+D^2}\left(\frac{\tan \omega x-\frac CD}{1+(\tan \omega x)\frac CD}\right)+\frac{AC+BD}{C^2+D^2}\\
   =\frac{AD-BC}{C^2+D^2}\left(\frac{D\sin \omega x-C\cos \omega x}{D\cos \omega x+C\sin \omega x}\right)+\frac{AC+BD}{C^2+D^2}\\
   =\frac{(AD-BC)(D\sin \omega x-C\cos \omega x)+(AC+BD)(D\cos \omega x+C\sin \omega x)}{(C^2+D^2)(C\sin \omega x+D\cos \omega x)}\\
   =\frac{(AD^2-BCD+AC^2+BCD)\sin \omega x+(-ACD+BC^2+ACD+BD^2)\cos \omega x}{(C^2+D^2)(C\sin \omega x+D\cos \omega x)}\\
   =\frac{A\sin \omega x+B\cos \omega x}{C\sin \omega x+D\cos \omega x}.$$
A: The answer is yes.
Claim 1 : If $D=0$, then
$$\dfrac{y_1}{y_2}=-\frac BC \tan\bigg(\omega x -\frac{\pi}{2}\bigg) + \frac AC$$
Claim 2 : If $D\not=0$, then $$\frac {y_1}{y_2}=\frac{DA-BC}{C^2+D^2}\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)+\frac{CA+BD}{C^2+D^2}$$

Claim 1 : If $D=0$, then
$$\dfrac{y_1}{y_2}=-\frac BC \tan\bigg(\omega x - \frac{\pi}{2}\bigg) + \frac AC$$
Proof :
One has
$$\frac{y_1}{y_2}=\frac BC\cdot\frac{\cos(\omega x)}{\sin(\omega x)}+\frac AC=-\frac BC \tan\bigg(\omega x - \frac{\pi}{2}\bigg) + \frac AC\ .\quad\blacksquare$$

Claim 2 : If $D\not=0$, then $$\frac {y_1}{y_2}=\frac{DA-BC}{C^2+D^2}\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)+\frac{CA+BD}{C^2+D^2}$$
Proof :
If $A=\dfrac{BC}{D}$, then
$$\frac{y_1}{y_2}=\frac{B(C\sin(\omega x)+D\cos(\omega x))}{D(C\sin(\omega x)+D\cos(\omega x))}=\frac BD$$
In the following, $A\not=\dfrac{BC}{D}$.
For $x$ such that $\cos(\omega x)\not=0$, one has
$$\frac{y_1}{y_2}=\frac{ A\dfrac{\sin(\omega x)}{\cos(\omega x)} + B}{C\dfrac{\sin(\omega x)}{\cos(\omega x)} + D}=\frac{A\tan(\omega x)+B}{C\tan(\omega x)+D}=\frac{\dfrac AD\tan(\omega x)+\dfrac BD}{\dfrac CD\tan(\omega x)+1}$$
Now,
$$\begin{align}G\bigg(\frac{y_1}{y_2}+F\bigg)&=\frac{G\dfrac AD\tan(\omega x)+G\dfrac BD}{\dfrac CD\tan(\omega x)+1}+GF
\\\\&=\frac{G\dfrac AD\tan(\omega x)+G\dfrac BD+GF\bigg(\dfrac CD\tan(\omega x)+1\bigg)}{\dfrac CD\tan(\omega x)+1}
\\\\&=\frac{G\bigg(\dfrac AD+\dfrac{CF}{D}\bigg)\tan(\omega x)-G\bigg(-\dfrac BD-F\bigg)}{1+\dfrac CD\tan(\omega x)}\end{align}$$
Solving the system
$$\begin{cases}G\bigg(\dfrac AD+\dfrac{CF}{D}\bigg)=1
\\G\bigg(-\dfrac BD-F\bigg)=\dfrac CD\end{cases}$$
for $F$ and $G$, one has
$$G=\frac{C^2+D^2}{DA-BC},\qquad F=\frac{-CA-BD}{C^2+D^2}$$
So, taking $(F,G)=\bigg(\dfrac{-CA-BD}{C^2+D^2},\dfrac{C^2+D^2}{DA-BC}\bigg)$, one has
$$G\bigg(\frac{y_1}{y_2}+F\bigg)=\frac{\tan(\omega x)-\dfrac CD}{1+\dfrac CD\tan(\omega x)}=\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)$$
so one gets
$$\frac {y_1}{y_2}=\frac 1G\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)-F,$$
i.e.
$$\frac {y_1}{y_2}=\frac{DA-BC}{C^2+D^2}\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)+\frac{CA+BD}{C^2+D^2}$$
which holds even for $x$ such that $\cos(\omega x)=0$ since for $x=\dfrac{1}{\omega}\bigg(\dfrac{\pi}{2}+n\pi\bigg)$ where $n$ is an integer, one has
$$\begin{align}\text{RHS}&=\frac{DA-BC}{C^2+D^2}\tan\bigg(\dfrac{\pi}{2}+n\pi-\arctan\bigg(\frac CD\bigg)\bigg)+\frac{CA+BD}{C^2+D^2}
\\\\&=\frac{DA-BC}{C^2+D^2}\cdot\frac{\cos\bigg(\arctan\bigg(\dfrac CD\bigg)\bigg)}{\sin\bigg(\arctan\bigg(\dfrac CD\bigg)\bigg)}+\frac{CA+BD}{C^2+D^2}
\\\\&=\frac{DA-BC}{C^2+D^2}\cdot\frac{1}{\dfrac CD}+\frac{CA+BD}{C^2+D^2}
\\\\&=\dfrac AC
\\\\&=\dfrac{y_1}{y_2}\end{align}$$
Therefore, one finally can say that if $D\not=0$, then
$$\frac {y_1}{y_2}=\frac{DA-BC}{C^2+D^2}\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)+\frac{CA+BD}{C^2+D^2}\ .\quad\blacksquare$$

Using this, your example can be written as
$$\frac{12 \sin (x) + 3 \cos(x)}{  4 \sin(x) + 2 \cos(x)}=\frac{3}{5}\tan\bigg(x-\arctan(2)\bigg)+\frac{27}{10}$$
A: Using $t=\omega x$ and assuming that all $(a,b,c,d)$ are non-zero, consider the function
$$F=(a \sin(t)+b \cos(t))-(c \sin(t)+d \cos(t))\big[e \tan(t+h)+k \big]$$ and we want that $F=0~~\forall ~x$.
So, it must be true for small values of $x$ and then use Taylor series around $t=0$. Since  we search for only three variables $(e,h,k)$, we shall not need a lot of terms.
Writing $$F=\sum_{n=0}^p \alpha_n\, t^n + O(t^{p+1})$$ the first coefficients are
$$\alpha_0=b-d e \tan (h)-d k \tag 1$$
$$\alpha_1=a-c e \tan (h)-c k-d e \tan ^2(h)-d e\tag 2$$
$$\alpha_2=-\frac{b}{2}-c e \tan ^2(h)-c e-d e \tan ^3(h)-\frac{1}{2} d e \tan (h)+\frac{d k}{2} \tag 3$$ and all of them mut be zero.
Solve $(1)$ for $k$. Plug it in $(2)$ and $(3)$. From the "new" $(2)$, solve it for $e$. Plug it in the "new" $(3)$  to obtain
$$(b c-a d) (c+d \tan (h))=0 \implies \tan(h)=-\frac c d$$ which then gives
$$\color{blue}{e=\frac{a d-b c}{c^2+d^2}\qquad\qquad h=-\tan ^{-1}\left(\frac{c}{d}\right) \qquad \qquad k=\frac{a c+b d}{c^2+d^2}}$$
You can check that all other coefficients are equal to $0$.
Applied to your case
$$\frac{12 \sin (x) + 3 \cos(x)}{  4 \sin(x) + 2 \cos(x)}=\frac{3}{5} \tan \left(x-\tan ^{-1}(2)\right)+\frac{27}{10}$$
Edit
Two nice applications of the above
$$\frac d {dx} \Bigg[\frac{a \sin (x) + b \cos(x)}{  c \sin(x) + d \cos(x)}\Bigg]=e \sec ^2(x+h)$$
$$\int \frac{a \sin (x) + b \cos(x)}{  c \sin(x) + d \cos(x)}\,dx=k x-e \log (|\cos (x+h)|)$$
