sigma_g :A-->A
Sigma_g (a) = g.a
this is a group G acting on the set A. for each fixed g in G we get a map sigma_g
i) for each fixed g in G , sigma_g is a permutation of A
ii) the map from G to S_A defined by g--> sigma_g is a homomorphism

this is a group action from the group G acting on the Set A .
sigma_g = g.a, is a permutation of A. like hmm

Skolem's paradox:
1. ZF has a countable model M. [Downward Löwenheim–Skolem]
2. The axiom of infinity guarantees that $\Bbb N$ is an object in M.
3. The axiom of powerset guarantees that $\mathscr P(\Bbb N)$ is an object in M.
4. Cantor's theorem proves that there is no surjection $\Bbb N \to \mathscr P(\Bbb N)$.
5. Since $\Bbb N$ is countable and our universe is countable, there must be a surjection $\Bbb N \to \mathscr P(\Bbb N)$.

1 0
0 i
is invertible

but none of
1 0
0 0
or
0 0
0 1
is invertible

no, this just proves that P cannot be
1 0
0 i

Show that $$\dfrac{\sec^2 \theta-\tan\theta }{\sec^2 \theta+\tan\theta}$$ lies between $$\dfrac{1}{3} \quad \mathbb{and} 3$$
Attempt:
I used $1+\tan^2\theta = \sec^2\theta$ to find that the denominator is always greater than one. Then I framed an inequality and eliminated the denominator. FInally, I am left with:
$$\tan^2\theta-\tan\theta+1>0$$ . How do I proceed? Any hints?

Please don't answer if it's unnecessary busy work for you/ wastes your time/ etc etc.