<s>[INST] rearranging a trig equation this is a problem that im not 100% can be done as i derived the equation myself but please help if you can and if it cant be done let me know:
i need the end product to be $\tan(\theta) = \ldots$
and for there to be no $\theta$ on the RHS.
equation is:
$$\frac{96.04 \cdot D^2}{\left(U \cdot \cos(\theta)\right)^2} = 2\left(U \cdot \sin(\theta)\right)^2 + (19.6 \cdot W)$$
where D, U and W are constants.
thanks so much in advance!!
------edit-------
i believe i owe you both an apology, it turns out i made a mistake earlier on in my equation so the one i posted was not possible. i only came to this conclusion after subbing in the data values i had into both answers and getting imaginary numbers back. so if i could be a pain, here is the updated (and verified) equation:
\begin{equation*}
\frac{D}{U\cos (\theta )}=\frac{U\sin (\theta )}{C}+( \frac{A+U^{2}\sin
^{2}(\theta )}{B}) ^{0.5}
\end{equation*}
once again im after $\tan(\theta)=$ etc with no ($\theta$) on the RHS
sorry for the mess around and thanks for helping :)
 [/INST] UPDATE (answer to the revised question). If we multiply the edited equation by $C\sqrt{B}U\cos \theta \neq 0$, we get
the equivalent equation
\begin{equation*}
\sqrt{B}CD=\sqrt{B}U^{2}\sin \theta \cos \theta +CU\left( A+U^{2}\sin
^{2}\theta \right) ^{1/2}\cos \theta .
\end{equation*}
Using the definition of $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and
the fundamental trigonometric identity $\sin ^{2}\theta +\cos ^{2}\theta =1$
, we are able to express $\sin \theta $ and $\cos \theta $, as well as $\sin
\theta \cos \theta $, in terms of $t=\tan \theta $. A possible derivation is
as follows. From
\begin{equation*}
\tan ^{2}\theta =\frac{\sin ^{2}\theta }{\cos ^{2}\theta }=\frac{\sin
^{2}\theta }{1-\sin ^{2}\theta }=\frac{1-\cos ^{2}\theta }{\cos ^{2}\theta }=
\frac{1}{\cos ^{2}\theta }-1
\end{equation*}
we obtain successively
\begin{eqnarray*}
\cos ^{2}\theta  &=&\frac{1}{1+\tan ^{2}\theta } \\
&\Rightarrow &\cos \theta =\pm \frac{1}{\sqrt{1+\tan ^{2}\theta }}=\pm \frac{
1}{\sqrt{1+t^{2}}} \\
&& \\
\sin ^{2}\theta  &=&\left( 1-\sin ^{2}\theta \right) \tan ^{2}\theta
\Leftrightarrow \sin ^{2}\theta =\frac{\tan ^{2}\theta }{1+\tan ^{2}\theta }=
\frac{t^{2}}{1+t^{2}} \\
&\Rightarrow &\sin \theta =\pm \frac{\tan \theta }{\sqrt{1+\tan ^{2}\theta }}
=\pm \frac{t}{\sqrt{1+t^{2}}}
\end{eqnarray*}
and
\begin{eqnarray*}
\sin \theta  &=&\cos \theta \tan \theta  \\
&\Leftrightarrow &\sin \theta \cos \theta =\cos ^{2}\theta \tan \theta =
\frac{\tan \theta }{1+\tan ^{2}\theta }=\frac{t}{1+t^{2}}.
\end{eqnarray*}
Substituting these expressions in the first equation yields
\begin{eqnarray*}
\sqrt{B}CD &=&\frac{\sqrt{B}U^{2}t}{1+t^{2}}+CU\left( A+\frac{U^{2}t^{2}}{
1+t^{2}}\right) ^{1/2}\left( \pm \frac{1}{\sqrt{1+t^{2}}}\right)  \\
\sqrt{B}CD\left( 1+t^{2}\right) -\sqrt{B}U^{2}t &=&\pm CU\left( A+\left(
A+U^{2}\right) t^{2}\right) ^{1/2}
\end{eqnarray*}
Squaring now both sides and rearranging the terms we obtain the following
quartic equation in $t$
\begin{equation*}
at^{4}+bt^{3}+ct^{2}+dt+e=0
\end{equation*}
whose coefficients are
\begin{eqnarray*}
a &=&BC^{2}D^{2} \\
b &=&-2BCDU^{2} \\
c &=&2BC^{2}D^{2}-C^{2}U^{4}+BU^{4}-AC^{2}U^{2} \\
d &=&-2BCDU^{2} \\
e &=&-2BCDU^{2}-AC^{2}U^{2}+BC^{2}D^{2}.
\end{eqnarray*}

unfortunately this is the simplest form i can get it to. is it still possible to do or am i wasting my time?

This equation (after depressed) although solvable algebraically by an auxiliary cubic equation has huge solutions (see Wikipedia entry Ferrari's solution). 

Multiplying your equation by $U^{2}\cos ^{2}\theta\ne 0 $ we get the equivalent
equation
\begin{equation*}
2U^{4}\sin ^{2}\theta \cos ^{2}\theta +19.6WU^{2}\cos ^{2}\theta
-96.04D^{2}=0.
\end{equation*}
Using the identity $\sin ^{2}\theta =1-\cos ^{2}\theta $ and rearranging the
terms we obtain the following quadratic equation in $y=\cos ^{2}\theta $:
\begin{equation*}
ay^{2}+by+c=0\Leftrightarrow y=\frac{1}{2a}\left( -b\pm \sqrt{b^{2}-4ac}
\right) ,
\end{equation*}
whose coefficients are
\begin{equation*}
a=-2U^{4},\quad b=2U^{4}+19.6WU^{2},\quad c=-96.04D^{2}.
\end{equation*}
To get the result we just need to express $\tan \theta $ in terms of $y$
\begin{equation*}
\sin ^{2}\theta +\cos ^{2}\theta =1\Leftrightarrow \tan ^{2}\theta +1=\frac{1
}{\cos ^{2}\theta }=\frac{1}{y}\Leftrightarrow \tan \theta =\pm \sqrt{\frac{1}{y}-1}.
\end{equation*}
</s>