Does the fact that $\tan^2(x)={\sin^2x\over \cos^2x}$ on $(-{\pi\over 2},{\pi\over 2})$ means $\{\tan^2(x),\cos^2x,\sin^2x\}$ are linearly dependent? Does the fact that $\tan^2(x)={\sin^2x\over \cos^2x}$ on $(-{\pi\over 2},{\pi\over 2})$ mean $\{\tan,\cos,\sin\}$ is linearly dependent?
I am not sure what to say. It does mean that. Doesn't it? What am I missing here to be asked such a seemingly trivial question? 
 A: No, this does not mean your set is L.D, I think what this exercise is about is the following:
Consider a linear combination giving $0$:
$$A\tan^2x+B\cos^2x+C\sin^2x=0\tag 1$$
One might be tempted to say "Let $A=1,B=0,C=-\frac {1}{\cos^2x}$ and we have our nontrivial combination of $A,B,C$ giving us the $0$ function!" but the thing is that $A,B,C$ should be numbers, not functions (this is, assuming that the vector space you're talking is $(\Bbb R^{(-\frac \pi 2,\frac \pi 2)},\Bbb R,\cdot,+)$).
Remark
To say that a set of $n$ functions is linearly independent means that if we take a linear combination giving the $0$ function:
$$a_1f_1(t)+a_2f_2(t)+...+a_nf_n(t)=0$$
It must be that $a_1=...=a_n=0$ for all $t$.
Going back to our equation (1): Let's evaluate it at $x=0$, we then have 
$$A\tan^2(0)+B\cos^2(0)+C\sin^2(0)=B\cos^2(0)=B=0$$
Thus $B=0$, now, if $x\neq 0$ we have that
$$A\tan^2x+C\sin^2x=0\iff A\sec^2x+C=0\iff A\sec^2x=-C$$
Now, we now that $sec^2x$ is not constant, so the only way it can equal a constant for all $x$ is that $A=0$. Looking back at (1), we deduce that $C$ must be $0$ aswell. So your set is $\text{L.I}$.
A: Hint: Suppose $a\tan^2 x + b\cos^2 x + c\sin^2 x \equiv 0$ on $(-\pi/2,\pi/2).$ Now one of those functions is unbounded on this interval, so ...
A: The Wronksian of these functions is given by:
$$W(x) = \det \left( \begin{matrix} \sin x & \cos x & \tan x \\ \cos x & -\sin x & 1 + \tan^2 x \\ -\sin x & -\cos x & \tan x(1 + \tan^2 x) \end{matrix} \right)$$
It simplifies to: 
$$W(x) = -\tan x(1 + 2\sec^2 x)$$
This is not identically zero on $(-\pi/2,\pi/2)$
