Product of tangents I was able to reduce an equation I have to:
$$f(t) = \tan(\mu) \tan(\nu) - C = 0$$
where $\mu, \nu$ are linear functions of t and $C$ is a constant.


*

*Are there any identities for the product of tangents?

*Is there a way to solve this equation analytically?

 A: Using the identity pointed out by J.M., we can transform the equation into 
$$\cos(\mu-\nu)-\cos(\mu+\nu) = C(\cos(\mu-\nu)+\cos(\mu+\nu)) \tag1$$
or better yet, 
$$(1-C)\cos(\mu-\nu) = (1+C) \cos(\mu+\nu) \tag2$$ 
Here both $\mu-\nu$ and $\mu+\nu$ are linear functions of $t$, which, as I infer from a comment by the OP, means linear and not affine. Assuming neither $\mu-\nu$ nor $\mu+\nu$ happen to be identically zero (which is an easy case), we can make (2) more compact:
$$\cos x = A\cos bx \tag3$$ 
where $A=(1+C)/(1-C)$ and $b=(\mu+\nu)/(\mu-\nu)$, which is a nonzero constant. There are two cases: 
$b$ is rational, $b=m/n$. Then both sides of (3) are periodic functions with period $2\pi n$. Let a numerical routine find all roots of (3) on $[0,2\pi n]$, and write down all of them by periodicity. If both $m$ and $n$ are very small, an explicit solution may exist, but otherwise not. 
$b$ is irrational. The function $\cos x-A\cos bx$ is not periodic, but it's quasiperiodic. This means that the roots of (3) come arbitrarily close to forming a periodic pattern, but they never do. Here is the graph of $\cos x-\pi \cos ex$, for example:
 
All I can add is an easy observation: if $A\ll 1$ or $A\gg 1$, the roots of (3) are close to the zeroes of the wave with greater magnitude. 
I wonder if (3) has a name...
A: This is one of my own tangent half-angle formulas (or "formulae", if you like), and its in a paper I'm writing on Villarceau's theorem:
$$
\begin{align}
\cos\gamma & =\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta} \\[10pt]
\text{and }\sin\gamma & = \frac{\sin\alpha\sin\beta}{1+\cos\alpha\cos\beta} \\[10pt]
\text{if and only if } \tan\frac\gamma2 & = \tan\frac\alpha2\cdot\tan\frac\beta2.
\end{align}
$$
A: $$
\tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta},
$$
$$
\tan(\alpha+\beta+\gamma) = \frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\alpha\tan\gamma-\tan\beta\tan\gamma},
$$
etc.
Each of these involves products of tangents.
