Solving for $\theta$ in trigonometric equations Please can someone help transpose the following trigonometric equations to make $m_r$ or $\tan\theta_r$ the subject ($m_t$ and $m_r$ are line gradients):
Solve each of the following equations, separately, for $\theta_r$ in terms of $m_t$:
$$m_t=\tan\theta_t=\frac{e-\cos\theta_r}{\sin\theta_r}=e\csc\theta_r-\cot\theta_r\tag{1}$$
$$m_t=\tan\theta_t=\frac{e^2\cos^2\theta_r-\cos^2\theta_r+\sin^2\theta_r}{2\sin\theta_r\cos\theta_r}=e^2\frac{1}{2}\cot\theta_r-\cot2\theta_r
\tag{2}$$
These are two independent equations, not a system of equations. For both equations I want to be able to solve:
$$m_r=\tan\theta_r=?$$
I've spent many hours trying but don't have the knowledge :(
Thanks in advance,
 A: You may first "reduce" to $\frac{e^2\cos^2\theta-2\cos^2\theta+1}{2\cos\theta}=e-\cos\theta$
From here you derive the equation (some work left for you): $e^2\cos^2\theta-2e\cos\theta+1=0$
Solve for $\theta$ to get $\theta= arcos(1/e)$
A: As @Cameron Williams has pointed out, you can write both of these as quadratic equations of a trigonometric function of $\theta_r.$

For equation (1), make the substitution $\cos\theta_r=\pm\sqrt{1-\sin^2\theta_r}$ and multiply through by $\sin\theta_r$ to get
$$e-m_t\sin\theta_r=\pm\sqrt{1-\sin^2\theta_r}.$$
Squaring both sides and setting equal to zero, we get
$$(m_t^2+1)\sin^2\theta_r-2em_t\sin\theta_r+(e^2-1)=0.$$
Solving for $\sin\theta_r$ using the Quadratic Formula gives
$$\sin\theta_r=\frac{em_t\pm\sqrt{m_t^2-e^2+1}}{m_t^2+1},$$
upon which you can use the inverse sine to extract $\theta_r.$

For equation (2), a similar trick will work.  But as I suggested in the comments, I will use double angle formulas.  Once I get to the quadratic form, I will leave it up to you to solve.
First multiplying through by $2\sin\theta_r\cos\theta_r,$ and then using the identities 


*

*$\sin2\theta_r=2\sin\theta_r\cos\theta_r$

*$\cos2\theta_r=\cos^2\theta_r-\sin^2\theta_r,$ and

*$\cos^2\theta_r=\dfrac{1+\cos2\theta_r}{2}$


we get
$$m_t\sin2\theta_r=\frac{e^2}{2}(1+\cos2\theta_r)-\cos2\theta_r.$$
Now multiplying through by 2, once again using the Pythagorean Identity $\sin2\theta_r=\pm\sqrt{1-\cos^2 2\theta_r},$ and squaring both sides, we get
$$4m_t^2(1-\cos^2 2\theta_r)=e^4+2e^2(e^2-1)\cos2\theta_r+(e^2-1)^2\cos^2 2\theta_r,$$
which is a quadratic equation in $\cos 2\theta_r$ that I will leave you to solve. 
A: Ah, they are separate problems--
First one is simpler. Not given.
Second one with abbreviations: Hope you guess what is omitted.
$ E = e^2/2,  C = \cos 2 \theta, S =\sin 2 \theta $
$ E (1+ C)/S - C/S = m_t $
$ E + C ( E-1) = m_t  S $
$ (1-E) C/A + m S / A = E/A $, where again $ A = \sqrt{(1-E)^2 + m_t^2}$
Let $ (1-E)/A = s_u , m/A = c_u $
$ s_u c _{2 \theta} + c_u s_{2 \theta} $
....
$ 2 \theta_r = \sin^{-1} (E/A) -\cos^{-1} (m_t/A) $
A: Answer #2
This is if you interpret the OP's question as: "write each expression on the RHS in terms of $\tan\theta_\color{red}r,$."
For `expression' (1),
$$e\csc\theta_r-\cot\theta_r=\pm e\sqrt{1+\cot\theta_r}-\cot\theta_r=\pm e\sqrt{1+\dfrac{1}{\tan\theta_r}}-\dfrac{1}{\tan\theta_r}.$$
For 'expression' (2), use the double angle formula $\cot(2\theta_r)=\dfrac{1-\tan^2\theta_r}{2\tan\theta_r}$ to get
$$\frac{e^2}{2}\cot\theta_r-\cot 2\theta_r=\frac{e^2}{2\tan\theta_r}-\frac{1-\tan^2\theta_r}{2\tan\theta_r}=\frac{e^2-1+\tan^2\theta_r}{2\tan\theta_r}.$$
Written in terms of $m_r$, expression (1) is
$$\pm e\sqrt{1+\frac{1}{m_r}}-\frac{1}{m_r},$$
and expression (2) is
$$\frac{e^2-1+m_r^2}{2m_r}.$$
Does this answer your question?
A: The first equation is, using $m=m_t$ and $\theta=\theta_r$,
$$
m\sin\theta=e-\cos\theta
$$
Set $t=\tan(\theta/2)$, so you get
$$
1-t^2+2mt=e+et^2
$$
that becomes the standard quadratic
$$
(1+e)t^2-2mt+e-1=0
$$
Just observe that $\theta=\pi$ is not a solution unless $e=-1$, case that you can treat separately.
The second equation can be written
$$
\sin^2\theta-2m\sin\theta\cos\theta+(e^2-1)\cos^2\theta=0
$$
Note that $\cos\theta\ne0$, so this becomes
$$
\tan^2\theta-2m\tan\theta+e^2-1=0
$$
which is a quadratic in $\tan\theta$.
