If $\tan^{-1} \left(\frac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$, then prove $x^2=\sin(2\alpha)$

Question

If $\tan^{-1} \left(\dfrac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$ then prove that: $x^2= \sin (2\alpha) $

My Attempt: $$\tan^{-1} \left(\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) =\alpha$$ $$\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}=\tan (\alpha )$$ $$\dfrac {1+x^2-2\sqrt {1+x^2}.\sqrt {1-x^2}+ 1 - x^2}{1+x^2-1+x^2}=\tan (\alpha)$$ $$\dfrac {1-\sqrt {1+x^2}.\sqrt {1-x^2}}{x^2}=\tan (\alpha)$$

Answer 1

Continuing from where you stopped

We get

$$1-x^2\tan\alpha= \sqrt {1+x^2} .\sqrt {1-x^2}$$

Squaring both sides we get $$1+x^4\tan^2\alpha-2x^2\tan\alpha=1-x^4$$ $$x^4(1+\tan^2\alpha)=2x^2\tan\alpha$$ $$x^2=\frac {2\tan\alpha}{1+\tan^2\alpha}$$ Hence $$x^2=\sin(2\alpha)$$

Answer 2

Since \begin{align} \tan\alpha = \frac{\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}} =\frac{\text{opposite}}{\text{adjacent}} \end{align} then it follows \begin{align} \sin \alpha =& \frac{\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt{(\sqrt {1+x^2} - \sqrt {1-x^2})^2+(\sqrt {1+x^2} + \sqrt {1-x^2})^2}}\\ =&\ \frac{\sqrt {1+x^2} - \sqrt {1-x^2}}{2} = \frac{\text{opposite}}{\text{hypotenuse}} \end{align} and \begin{align} \cos\alpha = \frac{\sqrt {1+x^2} + \sqrt {1-x^2}}{2} = \frac{\text{adjacent}}{\text{hypotenuse}}. \end{align} Then it follows \begin{align} 2\sin\alpha \cos \alpha = \frac{(\sqrt {1+x^2} + \sqrt {1-x^2})(\sqrt {1+x^2} - \sqrt {1-x^2})}{2}=x^2. \end{align}

Answer 3

Let $A=\sqrt{1+x^2}$ and $B=\sqrt{1-x^2}$, then $$\tan(\alpha) = \frac{A-B}{A+B}, \quad A^2+B^2=2,\quad A^2-B^2 = 2x^2$$ and also note $$\tan(a) = \underbrace{\frac{\sin(2a)}{2\sin(a)\cos(a)}}_{=\,1} \cdot \frac{\sin(a)}{\cos(a)}= \sin(2a)\cdot\frac{1}{2}\sec^2(a) = \sin(2a) \frac{1}{2} \left(1+\tan^2(a)\frac{}{}\right)$$ hence, \begin{align} \sin(2a) &= \frac{2 \tan(a)}{1+\tan^2(a)} = \frac{2 \tan(a)}{1+\tan^2(a)} \cdot \frac{\big(A+B\big)^2}{\big(A+B\big)^2}\\ &= \frac{2 (A+B)(A-B)}{(A+B)^2 + (A-B)^2} = \frac{2 (A^2-B^2)}{2A^2+2B^2} = x^2 \end{align}

Answer 4

As for real $\alpha, x^2\le1$

WLOG $x^2=\cos2y\implies0\le2y\le\dfrac\pi2\implies\cos y,\sin y\ge0$

Using $\cos2y=1-2\sin^2y=2\cos^2y-1,$

$$\tan\alpha=\dfrac{\cos y-\sin y}{\cos y+\sin y}=\tan\left(\dfrac\pi4-y\right)$$

$\implies\alpha=m\pi+\dfrac\pi4-y$ where $m$ is any integer

$x^2=\cos2y=\cos\left(2m\pi+\dfrac\pi2-2\alpha\right)=?$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that \begin{align} \pars{{p \over q} = z \implies \left\{\begin{array}{lcl} \ds{p + q \over q} & \ds{=} & \ds{z + 1} \\[1mm] \mbox{and}&& \\[1mm] \ds{p - q \over q} & \ds{=} & \ds{z - 1} \end{array}\right.} \implies \bbx{{p \over q} = z \implies {p + q \over p - q} = {z + 1 \over z - 1}}\label{1}\tag{1} \end{align}

Then, \begin{align} &\arctan\pars{\root{1 + x^{2}} - \root{1 - x^{2}} \over \root{1 + x^{2}} + \root{1 - x^{2}}} = \alpha \implies {\root{1 + x^{2}} - \root{1 - x^{2}} \over \root{1 + x^{2}} + \root{1 - x^{2}}} = \tan\pars{\alpha} \end{align}

With the identity \eqref{1}:

\begin{align} &{2\root{1 + x^{2}} \over -2\root{1 - x^{2}}} = {\tan\pars{\alpha} + 1 \over \tan\pars{\alpha} - 1} \implies {1 + x^{2} \over 1 - x^{2}} = \bracks{\tan\pars{\alpha} + 1 \over \tan\pars{\alpha} - 1}^{2} \\[5mm] \stackrel{\mrm{see}\ \eqref{1}}{\implies}\,\,\ & {2 \over 2x^{2}} = {\braces{\bracks{\tan\pars{\alpha} + 1}/ \bracks{\tan\pars{\alpha} - 1}}^{\,2} + 1 \over \braces{\bracks{\tan\pars{\alpha} + 1}/ \bracks{\tan\pars{\alpha} - 1}}^{\,2} - 1} = {2\tan^{2}\pars{\alpha} + 2 \over 4\tan\pars{\alpha}} \\[5mm] \implies &\ x^{2} = {2\tan\pars{\alpha} \over \tan^{2}\pars{\alpha} + 1} = {2\tan\pars{\alpha} \over\sec^{2}\pars{\alpha}} = 2\sin\pars{\alpha}\cos\pars{\alpha} \implies \bbx{x^{2} = \sin\pars{2\alpha}} \end{align}