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I have messed up my memory about trig identities or even worse (which is the most plausible), never known all of them (and thus could not get the right keywords to find if this is answered).

Given

$$\frac{\tan(a)}{\tan(b)} = N$$

$a$ is always within bounds $(0, \pi/2)$, thus not running around the unit circle
$b$ is always within bounds $(0, \pi/2)$, thus not running around the unit circle
always $N > 0$

Image of two triangles

Is there a way to express the ratio between angles themselves in form of

$$\frac{a}{b} = xN$$

or some

$$\frac{a}{b} = f(x,N)$$

where $x$ (and/or $f$) is the magic sugar?

Edit:
I'd like to find a general (if possible) way to express how the angular diameter (which itself is angle $[rad]$) changes when distance is changed to the object (which is considered to be plane, not sphere) by multiplier $N$. The object size does not change.

Thank you!

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    The ratio between $\tan x$ and $\tan kx$, where $k$ is a constant of proportionality, is not in general a function only of $k$, but of $x$ as well, so I suspect what you search for does not exist—at least, not the way you want it. – 2017-02-02

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No. $\tan(a)/\tan(b)$ is not a function of $a/b,$ nor is $a/b$ a function of $\tan(a)/\tan(b).$

For instance if $a = \pi/3$ and $b=\pi/6$ then $a/b = 2$ and $\tan(a)/\tan(b) = 3.$ However if $a=\pi/4$ and $b=\pi/8$ then $a/b=2,$ but $\tan(a)/\tan(b) = 1+\sqrt{2}$. So the ratio $\tan(a)/\tan(b)$ is not determined by $a/b.$

Likewise, when $\tan(a)/\tan(b)=3,$ we have seen one possible solution is that $a=\pi/3$ and $b=\pi/6$ so that $a/b=2.$ However, if we let $\tan(b)=1$ so that $b=\pi/4$ and let $\tan(a) = 3$ so that $a=\arctan(3),$ then we still have $\tan(a)/\tan(b) = 3$ but $$a/b= \frac{\arctan(3)}{\pi/4} \approx 1.59 \ne 2.$$

So $a/b$ is not determined by $\tan(a)/\tan(b).$

Therefore each ratio $\tan(a)/\tan(b)$ corresponds to many possible ratios $a/b$ and vice versa. Thus you won't find a simple relationship to solve for one in terms of the other.

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    Thank you. That's what I thought, because I was running circles in $arctan$ and what not ratios. However updated the question, showing *practical* reason for this question. – 2017-02-02