Prove that $\tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac 12 \cos^{-1}x^2$

Let the above expression be equal to $$\phi$$ $$\frac{\tan \phi +1}{\tan \phi-1}=\sqrt{\frac{1+x^2}{1-x^2}}$$ $$\frac{1+\tan^2\phi +2\tan \phi}{1+\tan^2 \phi-2\tan \phi}=\frac{1+x^2}{1-x^2}$$

$$\frac{1+\tan^2\phi}{2\tan \phi }=\frac{1}{x^2}$$ $$\sin 2\phi=x^2$$ $$\phi=\frac{\pi}{4}-\frac 12 \cos^{-1}x^2$$

Where am I going wrong?

• arc sin (y) + arc cos (y) =$\pi/2$ for all $y\in [-1,1]$
– Koro
Jul 18, 2020 at 9:00
• @Koro please check the edit Jul 18, 2020 at 9:16
• Or 9 or 10 doesn't change if it's a good question. +1 Jul 18, 2020 at 9:59

4 Answers

Because for $$x\neq0$$ and $$-1\leq x\leq1$$ easy to see that: $$0<\frac{\pi}{4}+\frac 12 \cos^{-1}x^2<\frac{\pi}{2}$$ and we obtain: $$\tan\left(\frac{\pi}{4}+\frac 12 \cos^{-1}x^2\right)=\frac{1+\tan\frac{1}{2}\arccos{x^2}}{1-\tan\frac{1}{2}\arccos{x^2}}=$$ $$=\frac{\cos\frac{1}{2}\arccos{x^2}+\sin\frac{1}{2}\arccos{x^2}}{\cos\frac{1}{2}\arccos{x^2}-\sin\frac{1}{2}\arccos{x^2}}=\frac{\sqrt{\frac{1+x^2}{2}}+\sqrt{\frac{1-x^2}{2}}}{\sqrt{\frac{1+x^2}{2}}-\sqrt{\frac{1-x^2}{2}}}=\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}.$$

Your mistake in the last line.

Indeed, since $$\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}>1$$ and from here $$\frac{\pi}{4}<\phi<\frac{\pi}{2},$$ we obtain: $$2\phi=\pi-\arcsin{x^2}=\frac{\pi}{2}+\arccos{x^2}.$$

• But what went wrong in my proof? Jul 18, 2020 at 9:51
• Also $0\le \cos^{-1}x^2 \le \frac {\pi}{2} \implies \frac{\pi}{4}\le \frac{\pi}{4} +\frac 12 \cos^{-1} x^2 \le \frac{\pi}{2}$ right? Jul 18, 2020 at 9:54
• @Aditya I added something. See now. Jul 18, 2020 at 10:02
• I am still not sure about the restriction on $\phi$ Jul 18, 2020 at 10:09
• @Aditya I added something again... Jul 18, 2020 at 10:15

A bit late answer but I thought worth mentioning it.

First note that we can substitute $$y=x^2$$ and consider $$0. Furthermore, the argument of $$\arctan$$ can be simplified as follows:

$$\frac{\sqrt{1+y}+\sqrt{1-y}}{\sqrt{1+y}-\sqrt{1-y}}=\frac{1+\sqrt{1-y^2}}{y}$$

Now, setting $$y = \cos t$$ for $$t \in \left[0,\frac{\pi}2\right)$$, to show is only

$$\arctan \frac{1+\sin t}{\cos t} = \frac{\pi}{4}+\frac t2$$

At this point half-angle formulas come into mind:

$$\frac{1+\sin t}{\cos t} = \frac{(\cos \frac t2 + \sin \frac t2)^2}{\cos^2 \frac t2 - \sin^2 \frac t2} = \frac{\cos \frac t2 + \sin \frac t2}{\cos \frac t2 - \sin \frac t2}$$ $$= \frac{1+\tan \frac t2}{1-\tan \frac t2} = \tan\left(\frac{\pi}{4}+\frac t2\right)$$.

Done.

Domain of $$\tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$$ is $$x \in (-1,1]$$, nothing wrong with that.

But range of the above term(argument of $$\arctan$$) is $$(1,\infty)$$ so this means, when you assume it be $$\phi$$, it is restricted to the interval $$\left[ \frac{\pi}{4},\frac{\pi}{2}\right]$$

This creates a problem in the last line, because $$\sin^{-1}(x^2)$$ should be in its principal range and $$\frac{\pi}{2}\le 2\phi \le \pi$$

Edit: to find the domain of the argument, the easiest way, divide both sides by $$\sqrt{1+x^2}$$ and then substitute $$x^2=\cos(2\theta)$$, $$\theta \in \left(0,\frac{\pi}{4}\right)$$, to get

$$\frac{1+\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}}{1-\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}}$$

and then use the identity, $$\tan \theta=\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}$$ $$\frac{1+\tan \theta}{1-\tan \theta}=\tan \left(\frac{\pi}{4}+ \theta\right)$$ Now, $$\theta \in \left(0,\frac{\pi}{4}\right)$$ so, $$\tan \left(\frac{\pi}{4}+\theta\right) \in [1, \infty)$$

Which is ironically your question...

• Can you please explain how you range for arctan, because $x\not >1$ Jul 18, 2020 at 10:07
• Either you plot the graph, or use calculus to find the domain of the argument of arctan. @Aditya Jul 18, 2020 at 10:12
• @Aditya, looks like you got another way to solve it Jul 18, 2020 at 10:25

Your mistake

$$\sin y=a\stackrel{\text{to}}{\longrightarrow}\sin^{-1}(\sin y)=\begin{cases}2n\pi+y&y\in\text{I, IV quadrant}\\(2n-1)\pi-y&y\in\text{II, III quadrant}\end{cases}=\sin^{-1}a$$