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$ ?

In terms of sine and cosine, the expression you have between the $1/3$ and the $3$ is $\cos^2 \theta /(1+\sin^2 \theta).$ One can certainly show this is at most $1$, hence at most $3$ as your inequality says on one side. However the value is zero whenever cosine is zero, so that one cannot say the expression is at least $1/3$ as one half of your inequality claims.
