# 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 $\dfrac{2\cos 2\theta}{3(1+\sin^2 \theta)}\geq 0$ ?

• go on,you will have $2-4sin^2\theta \ge 0$ – chenbai Mar 1 '14 at 2:23
• @chenbai please, see my updated post – Soumitra Sen Mar 1 '14 at 2:33

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.

We can't show $$\dfrac{2\cos 2\theta}{3(1+\sin^2\theta)}\ge\iff\cos 2\theta\ge0$$ which can be $<0$ if $\displaystyle2n\pi+\frac\pi2<2\theta<2n\pi+\pi+\frac\pi2$ where $n$ is an integer

Why don't we adapt the following method:

$$\frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^2\theta}=\frac1{2\tan^2\theta+1}$$

Now, $$0\le\tan^2\theta\le\infty\iff0\le2\tan^2\theta\le\infty\iff1\le1+2\tan^2\theta\le\infty$$

$$\implies1\ge\frac1{1+2\tan^2\theta}\ge0\iff 0\le\frac1{1+2\tan^2\theta}\le1$$

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 $f(x)$

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 $\theta$?

• You write : "You cannot multiply through by $1+u$ because $1+u$ may be either positive or negative, depending on the value of $u$" but at the same time $u = \sin^2\theta$. Am I missing something ? Cheers. – Claude Leibovici Mar 1 '14 at 7:59