Solving $\tan\beta\sin\gamma-\tan\alpha\sec\beta\cos\gamma=b/a$, $\tan\alpha\tan\beta\sin\gamma+\sec\beta\cos\gamma=c/a$ for $\beta$ and $\gamma$ I am trying to solve the following tricky system of two trigonometric equations. $\alpha, a, b, c$ are all given (and $a$ is nonzero) so I am trying to solve for $\beta$ and $\gamma$.
$$\tan{\beta}\sin{\gamma}-\tan{\alpha}\sec{\beta}\cos{\gamma}=\frac{b}{a}$$
$$\tan{\alpha}\tan{\beta}\sin{\gamma}+\sec{\beta}\cos{\gamma}=\frac{c}{a}$$
If this is too difficult to solve by hand, are there any pieces of software/programming libraries that could help me solve this? I plugged it into woflramalpha and got a horrendous ~34 line equation for $\gamma$. I am hoping that a cleaner solution exists after some simplification.
 A: In order to see more symmetry, multiply both equations
$$
\left\{ 
\begin{aligned}
\tan\beta \sin\gamma - \tan\alpha \sec\beta \cos\gamma 
&= \frac{b}{a} \\[3pt]
\tan\alpha \tan\beta \sin\gamma + \sec\beta \cos\gamma 
&= \frac{c}{a} 
\end{aligned} 
\right. 
$$
by $\cos \alpha$ to get
$$
\left\{ 
\begin{aligned}
\cos\alpha \tan\beta \sin\gamma - \sin\alpha \sec\beta \cos\gamma 
&= \frac{b}{a} \\[3pt]
\sin\alpha \tan\beta \sin\gamma + \cos\alpha \sec\beta \cos\gamma 
&= \frac{c}{a} 
\end{aligned} 
\right. 
$$
which is of the form
$$
\left\{ 
\begin{aligned}
U \cos\alpha - V \sin\alpha 
&= \frac{b}{a} \\[3pt]
U \sin\alpha + V \cos\alpha 
&= \frac{c}{a} 
\end{aligned} 
\right. 
$$
for $U = \tan\beta \sin\gamma$ and $V = \sec\beta \cos\gamma$. This is rotation transformation of coordinates $(U, V)$ which we can invert:
$$
\left\{ 
\begin{aligned}
U &= \frac{b}{a} \cos\alpha + \frac{c}{a} \sin\alpha \\[3pt]
V &= -\frac{b}{a} \sin\alpha + \frac{c}{a} \cos\alpha 
\end{aligned} 
\right. 
$$
Now, we need to solve the equations for $(U, V)$ in terms of $(T, C) = (\tan\beta, \cos\gamma)$. Squaring gives us
$$
\left\{ 
\begin{aligned}
U^2 &= T^2 (1 - C^2) \\ 
V^2 &= (1 + T^2) C^2 
\end{aligned} 
\right. 
$$
This is a system of quadratic equations in $T^2$ and $C^2$, so it has at most $4$ solutions.
