Can someone explain this formula to me. It is a tapered hole I could not figure out how to do a couple of the symbols so I am inserting an image.
I am trying to figure out how to use this formula.  It does involve machining a tapered hole but I am at a loss on how to use it.picture of formula
Transcription of formula: 
$$
\Delta H = 0.1 - \frac{0.025}{\tan \frac{\theta}{2}}.
$$
Here is some more information.
more information
 A: John. 
I assume that this is for some sort of machining process, and I haven't much experience with machining. But let me try to make a guess or two. 
If your "tapered" hole had no taper, it'd be a cylinder, right? In that case, we'd say that the angle of taper is $\theta = 0$. If the sides tilted in just a little bit, we might say that the taper was $\theta = 1$ degree. If it looked like a "90 degree) countersink, we might say that the taper was $\theta = 45$ degrees (i.e., the taper is the slope of one side of the hole, compared to vertical). I could easily be wrong here: the taper might be the angle between the two opposite sides, so that for a 90-degree countersink, the taper would be $90$ degrees. The use of $\theta/2$ in your formula suggests to me that this latter interpretation might be the correct one, but without more information, I can't say. 
Let's suppose that the taper is $\theta = 10$ degrees. Then $\theta/2$ is $5$ degrees, and $\tan \frac{\theta}{2} \approx 0.0875$ (I got this answer using a calculator set to "degrees" mode!)
Then your formula says that 
$$
\Delta H 
= 0.1 - \frac{0.025}{\tan \frac{\theta}{2}} 
\approx 0.1 - \frac{0.025}{0.0875} 
\approx 0.1 - 0.2857 \approx -0.1857
$$
where $\approx$ means "is approximately equal to," because I suspect you don't want more than a few decimal digits of accuracy for machining. 
The symbol $\Delta$ is often used to mean "the change in", so this seems to say that the "change in H" should be $-0.1857$. I don't know what $H$ is in your setup, but this suggests reducing it by a modest amount. 
I'm sorry not to be able to be more confident in my answer, but to do so, I'd need a bit more context. 
