Would appreciate if someone can help me with this one.
The angle between the asymptotes of a hyperbola is $\dfrac{\pi}{3}$. Determine the eccentricity of the hyperbola.
Since I'm trying to self teach myself here, the only thing I could find was that the tangent of the angle between the asymptotes is $\dfrac{2ab}{a^2-b^2}$. I also know the formula for the eccentricity but I can't figure out how to find it out of that angle alone.
(Note: this answer shows how to work it out using the supplied formula for the tangent, so I'll keep it up, but there's a much simpler graphical method here.)
First off, there's an typo in your formula for eccentricity. It should be $\sqrt{(a^2 + b^2)/a^2}$.
With that out of the way, the key point to figuring this out is to realize that both the eccentricity and the angle really only depend on the ratio $b/a$, as they both measure how "squashed" the hyperbola is. So our job is now figuring out how to massage the formulas so they only depend on $b/a$, and not on $a$ or $b$ individually. For the tangent of the angle
$$\dfrac{2ab}{a^2-b^2} = \dfrac{1/a^2}{1/a^2}\cdot \dfrac{2ab}{a^2-b^2} = \dfrac{2ab/a^2}{a^2/a^2-b^2/a^2} = \dfrac{2(b/a)}{1-(b/a)^2}. $$
You can do something similar for the eccentricity formula.
Can you take it from there? Me, once I've rewritten the formulas in terms of $b/a$, I might introduce a new variable $k= b/a$ just to make the subsequent manipulations easier.
Consider the origin of axes situated in one of the foci. It is known (https://17calculus.com/conics/polar/) that, up to a rotation that doesn't change the eccentricity $e$, the polar equation of a hyperbola can be taken as:
$$\tag{1}p=\dfrac{k}{1-e \sin(\theta)}$$
where $e$ is its eccentricity.
(1) is undefined for values of $\theta$ such that:
$$\tag{2}1-e \sin(\theta)=0$$
which, precisely correspond to the directions of asymptotes (directions where "$r$ is infinite").
Let us define $\theta_0$ as the unique angular value in $(0, \dfrac{\pi}{2})$ such that :
$$\tag{3}1-e \sin(\theta_0)=0$$
(explanation : for a hyperbola, $e>1 \ \implies \ \tfrac{1}{e}<1$ ; and $\sin$ function is bijective from $(0, \dfrac{\pi}{2})$ to $(0,1)$.)
Now by hypothesis, we also have a zero value in (2) for $\theta_0+\dfrac{\pi}{3}$:
$$\tag{4}1-e \sin(\theta_0+\dfrac{\pi}{3})=0$$
Subtracting (3) from (4), we get :
$$\tag{4}\sin(\theta_0+\dfrac{\pi}{3})=\sin(\theta_0)$$
from wich we conclude that $\theta_0+\dfrac{\pi}{3}=\pi-\theta_0$, i.e.,
$$\tag{5}\theta_0=\dfrac{\pi}{3}.$$
Plugging (5) into (3) gives
$$e=\dfrac{2}{\sqrt{3}} \approx 1.155$$
Look at this picture I got from Wikipedia
If the angle between the asymptotes is $\pi/3$ then the angle of the triangle at $M$ (call it $\theta$) is half that, $\pi/6$. Then $$\dfrac{b}{a} = \tan{\theta} = \tan(\pi/6) = 1/\sqrt{3}.$$ The formula for eccentricity is $\sqrt{(a^2 + b^2)/a^2}$ (there was a typo in your comment). We can re-write this as $$\sqrt{\dfrac{a^2 + b^2}{a^2}} = \sqrt{\dfrac{a^2}{a^2} + \dfrac{b^2}{a^2}} = \sqrt{1 + \left(\dfrac{b}{a}\right)^2},$$ and just plug in the value of $b/a$.
(Note: The formula for the tangent of the asymptotes' angle that you found is just a thinly disguised version of the tangent double angle formula. The picture lets us see that we just need half of that angle to compute $b/a$ with our eyes instead of going through a lot of algebra.)
Using the properties of Hyperbola, we know that the angle between asymptotes is given by $\sec^{-1}(e).$
Thereby inverse of secant function with the eccentricity $e$ gives us the angle between asymptotes Thereby $e=\sec(\text{angle given})$
Angle given is $\frac{pi}3$ and thus $e=2$
