Essential singularities of $\frac{e^{\tan z}}{z^2 +1}$ I am supposed to find the nature of the essential singularities of the function $f(z)=\frac{e^{\tan z}}{z^2 +1}$. What I find is that the points where $f(z)$ has essential singularities are the zeroes of $\cos z$. So, the number of essential singularities of $f(z)$ is countable, and they are isolated. But my book says it has uncountable number of essential singularities. I am little confused. I must mention that the book is a local one written for competitive examinations and has significant number of wrong solutions.

For every zero os $\cos{z}$, your function has an essential singularity. There are infinitely many of them, as, $\forall k\in \mathbb{Z}, \cos{(\pi + 2k\pi)}=0$, but of course, there are not uncountably many! So, yes, the book is indeed wrong!
To make absolutelly sure, let's proof that that $\cos{(z)}$ has no zeros outside the real axis:
$$\cos{(z)} = \frac{e^{iz}+e^{-iz}}{2} = 0 \iff e^{iz}+e^{-iz} = 0$$
Let $z:=x+iy$, now:
$$e^{iz}+e^{-iz} = e^{i(x+iy)}+e^{-i(x+ìy)} = e^{-y+ix}+e^{y-ix} = e^{-y}e^{ix} +e^y e^{-ix}$$
This equals zero if, and only if:
$$e^{-y} e^{ix} = - e^y e^{-ix}$$, which is equivalent to:
$$e^{-y}(\cos{(x)} + i \sin{(x)}) = - e^y(\cos{(-x)} + i \sin{(-x)})$$
Considering the imaginary part of the expersion, it implies:
$$e^{-y} \sin{(x)} = -e^{y} \sin{(-x)}$$
So, as $\sin{(-x)}=-\sin{(x})$, $$e^{-y} = e^{y}$$, since exponentiation is injective, we conclude $y=-y$, therefore $y=0$
Your function has two more singularities at $z=i$ and $z=-i$. Those are not essential singularities, but order-$1$ poles.
