How can I rearrange a formula containing both cot and cosec? I want to rearrange the following:
$$\cot{\theta}-\csc{\theta}=\dfrac{2B\mu}{bmg}$$
to get $\theta$.
I am unsure how to reduce a cot and cosec sum. Are there any maths tricks or identities that I am missing?
 A: Note that 
$$\begin{align}
\cot \theta-\csc \theta &=-\frac{1-\cos \theta}{\sin \theta}\\\\
&=-\frac{2\sin^2(\theta/2)}{\sin(2\theta/2)}\\\\
&=-\frac{2\sin^2(\theta/2)}{2\sin (\theta/2)\cos (\theta/2)}\\\\
&=-\tan (\theta/2)
\end{align}$$
So, we have the identity
$$\bbox[5px,border:2px solid #C0A000]{\cot \theta-\csc \theta =-\tan (\theta/2)}$$
which can be directly applied to the problem of interest.
A: If $\cot \theta - \csc\theta = (\cos\theta - 1)/\sin\theta = k$ then
$$\frac{(\cos\theta-1)^2}{1-\cos^2\theta} = k^2$$
Thus
$$(1+k^2)\cos^2\theta - 2\cos\theta + (1 - k^2) = 0$$
and 
$$\cos\theta = 1 \text{ or } \frac{1-k^2}{1+k^2}$$
$\cos\theta = 1$ is ruled out by the original equation and thus
$$\theta = \arccos\left(\frac{1-k^2}{1+k^2}\right)$$
A: Use $\csc ^2\theta-\cot^2 \theta =1$ i.e. $(\csc \theta-\cot \theta)(\csc \theta+\cot \theta) =1$
So you get that $(\csc \theta+\cot \theta)=\frac{1}{\frac{2B\mu}{bmg}}=\frac{bmg}{2B\mu}$
Adding we get, $\csc \theta=\frac{\frac{2B\mu}{bmg}+\frac{bmg}{2B\mu}}{2}$ and $\cot \theta=\frac{\frac{bmg}{2B\mu}-\frac{2B\mu}{bmg}}{2}$
We can solve for $\theta$ from this easily.
A: $$\cot\theta - \csc \theta = k$$
$$\cot\theta \pm \sqrt{\cot^2 \theta+1} = k$$
$$\pm \sqrt{\cot^2 \theta+1} = k-\cot\theta$$
$$\cot^2 \theta+1 = (k-\cot\theta)^2$$
$$\cot^2 \theta+1 = k^2-2k\cot\theta+\cot^2\theta$$
$$1 = k^2-2k\cot\theta$$
$$\cot\theta = \frac{k^2-1}{2k}$$
