Three vectors are coplanar then they are linearly dependent. Is it true for the bottom? Suppose there are three vectors $\sin x,\cos x,\tan x$. Now I'm pretty sure that these lie on the same plane. But can I find any constant coefficients that satisfies $c_1\sin x + c_2\cos x + c_3\tan x =0$? I couldn't.  Now please rectify me if there is some way to prove that they are linearly dependent.  Or please explain if they aren't linearly dependent how they are lying on the same plane ($xy$ plane)     .
 A: There are no such constants (unless they're all $0$; of course). In fact,$$x=0\implies c_1\sin(x)+c_2\cos(x)+c_3\tan(x)=c_2$$and therefore $c_2=0$. On the other hand$$x=\frac\pi2\implies c_1\sin(x)+c_3\tan(x)=\frac{c_1}{\sqrt2}+c_3=0$$and therefore $c_3=-\frac{c_1}{\sqrt2}$. Finally, see what happens when $x=\frac\pi6$, in order to deduce that $c_1=c_3=0$.
A: Those 3 functions are linearly independent. Plugging $x=0$, you get that $c_2=0$. Now as $x \to \pi/2$, $\sin$ is bounded while $\tan$ is not, therefore $c_3=0$. Finally this imposes $c_1=0$ as $\sin$ is not always vanishing.
A: Suppose that $c_1 \sin x + c_2 \cos x + c_3 \tan x =0$  for all $x \in ( - \pi/2, \pi/2)$.
With $x=0$ we get $c_2=0$. Hence $ \sin x (c_1+\frac{c_3}{\cos x})=0$ for all $x \in ( - \pi/2, \pi/2).$.
If $x \ne 0$ we derive $c_1+\frac{c_3}{\cos x}=0$ .
It is your turn to show that $c_1=c_3=0$.
A: Beware: strong confusion between vectors and functions!
Three coplanar vectors in $\mathbb{R}^3$ are indeed not linearly independent...
But the example you are giving is not three vectors, these are three functions... This is completely different. 
And in that case, the three functions are clearly independent from each other, but these functions are not elements of $\mathbb{R}^3$
EDIT: side note: they are represented in the $xy$ plane, this is not to say they are elements of thus plane...
A: Let $x$ tend to $\pi\over 2$ form below. Then $\tan x$ grows up to $\infty$ while $\sin x$ and $\cos x$ are both bounded. This leads to $c_3=0$. Also let $x=0$ and $x=\dfrac{\pi}{2}$ which yields to $c_2=0$ and $c_1=0$ respectively arguing that these three functions are linearly independent.
A: The confusion here is in the word "coplanar".
The functions are not elements of the $xy$-plane. $\sin$, $\cos$, and $\tan$ live in an infinite-dimensional vector space. Even though we draw them as graphs in a plane, they are not elements of the plane, since the plane is just a set of points.
