<s>[INST] 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$ ?
 [/INST] You have
$$
\frac 1 3 \le \frac{1-u}{1+u} \le 3
$$
where $u = \sin^2\theta$.
You cannot multiply through by $1+u$ because $1+u$ may be either positive or negative, depending on the value of $u$.
If $\dfrac{1-u}{1+u}\le 3$ then $\dfrac{1-u}{1+u}-3\le0$, so $\dfrac{1-u}{1+u}-\dfrac{3(1+u)}{1+u}\le 0$, and that becomes $\dfrac{-2-4u}{1+u} \le 0$.  From that one gets $\dfrac{\frac 1 2 + u}{1+u}\ge0$.  This changes signs at $-1/2$ and at $-1$, and we have
$$
\frac{\frac 1 2 + u}{1+u} \begin{cases} >0 & \text{if }u>-1/2, & \\ <0 & \text{if }-1/2>u>-1, \\ > 0 & \text{if } -1>u.  \end{cases}
$$
Observe that you never have $-1>\sin^2\theta$, so you can ignore the third piece above.  Then you need to ask: If $-1/2>\sin^2\theta>-1$, then what is $\sin\theta$ and then what is $\theta$?
</s>