# Disproving that $\sec^{-1} x>\tan^{-1} x$ for all $x\geq1$

I was just experimenting with WolframAlpha today and noticed that the solution of the inequality $\sec^{-1} x\tan^{-1} x$ is $x\leq1$. I tried to prove the inequality for $x\geq1$ to be false.

My attempt:

Let $h(x)=\sec^{-1} x-\tan^{-1} x$. We have to prove that $h(x)<0 \forall x\geq1$. Note that $h(1)=-\pi/4$.

If $h(x)$ is a decreasing function, then QED already.
If $h(x)$ is an increasing function, then we note that $\lim _{x\to\infty}h(x)=\pi/2-\pi/2=0$. Hence, QED.

Both cases above are monotonic because the functions $\sec^{-1} x$ and $\tan^{-1} x$ are themselves monotonic.

Is my proof correct?

• Of course $h(x)$ can be either increasing or decreasing only, but I noticed that both cases satisfy the inequality, so I was too lazy to delve into the double derivative :P Dec 30 '17 at 2:18

You want to prove that $$\sec^{-1} x<\tan^{-1} x\;\;\;\forall x\geq1$$

To prove this, let $y=\sec^{-1}x$, $z=\tan^{-1}x$. $$x=\frac1{\cos y}=\frac{\sin z}{\cos z}\\\implies\cos z=\sin z\cos y$$ Since we are looking at $x\ge1$, this means that $\sin z\ge\cos z$, i.e. $z\in[\pi/4,\pi/2)$, and $y\in[0,\pi/2)$.

Now fix $y$. Then $$\cos z=\cos y \sin z\\\implies |\cos z|=|\cos y|\cdot|\sin z|<|\cos y|\\\implies\cos z<\cos y$$since both are positive. Since $\cos$ is decreasing on this interval, this implies that $y<z$, i.e. $$\sec^{-1}x<\tan^{-1}x$$

Both cases above are monotonic because the functions $\sec^{-1} x$ and $\tan^{-1} x$ are themselves monotonic.

The sum of two monotonic functions is not necessarily monotonic. Try out $f(x)=\sin x-x, \ g(x)=\cos x+x$ and you see that $f(x)+g(x)$ is not monotonic. Same applies to the difference of two monotonic functions.

Thus, you cannot establish the monotonicity of $h(x)=\sec^{-1} x-\tan^{-1} x$ that way in your proof.

• Thanks for your answer! But, it seems like both of your examples are indeed monotonic. The points where they have first derivative zero happens to be an inflexion point. Dec 30 '17 at 3:49
• @GaurangTandon If you want avoid this, you can construct functions like y=sin(x)+(x-k)^3, k>0... hmm and there shouldn't be such inflection points. Dec 30 '17 at 4:01
• Yes, that's better. Dec 30 '17 at 4:39