How can I prove the trigonometric Problem? How can I show  the following trigonometric problem :
$$\frac{1}{3}\leq \frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^2    
    \theta}\leq 3$$
I have tried in the following way :
$$
\begin{align}
& \phantom{\Rightarrow}
\frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^2
    \theta}-\frac{1}{3} \\[8pt]
& \Rightarrow  \frac{1}{\sec^2\theta+\tan^2
    \theta}-\frac{1}{3} \\[8pt]
& \Rightarrow  \frac{\cos^2\theta}{1+\sin^2
    \theta}-\frac{1}{3} \\[8pt]
& \Rightarrow  \frac{1-\sin^2\theta}{1+\sin^2
    \theta}-\frac{1}{3} \\[8pt]
& \Rightarrow  \frac{2-4\sin^2\theta}{3(1+\sin^2
    \theta)} \\[8pt]
& \Rightarrow  \frac{2\cos 2\theta}{3(1+\sin^2
    \theta)}
\end{align}
$$
How can I show $ \dfrac{2\cos 2\theta}{3(1+\sin^2    
    \theta)}\geq 0$ ?

I think you need to find  $$\frac13\leq \frac{\sec^2\theta-\tan\theta}{\sec^2\theta+\tan\theta}\leq 3$$
Setting $\tan\theta=x,$ we get $$\frac{\sec^2\theta-\tan\theta}{\sec^2\theta+\tan\theta}=\frac{x^2+1-x}{x^2+1+x}$$
Let $\displaystyle\frac{x^2+1-x}{x^2+1+x}=y$
$\displaystyle\iff (y-1)x^2+(y+1)x+y-1=0$ which is a Quadratic Equation if $x$
As $x$ is real, the discriminant must be $\ge0$  
$\displaystyle\implies (y+1)^2\ge4\cdot(y-1)^2\iff 3y^2-10y+3\le0\iff (3y-1)(y-3)\le0 $
As  the product of two multiplicand is $\le0$
So, we need $\displaystyle(i)3y-1\ge0\iff y\ge\frac13$ and $\displaystyle y-3\le0\iff y\le3\implies\frac13\le y\le3$
or $(ii)3y-1\le0\iff y\le\frac13$ and $y-3\ge0\iff y\ge3$ which is impossible

Alternatively, 
I leave for you the Second Derivative Test of $$f(x)=\frac{x^2+1-x}{x^2+1+x}$$ to find the extreme values of $f(x)$
