Question 11b8e

Nov 13, 2017

Explanation:

${\tan}^{-} 1 \left(\frac{\sqrt{1 + {x}^{2}} - \sqrt{1 - {x}^{2}}}{\sqrt{1 + {x}^{2}} + \sqrt{1 - {x}^{2}}}\right) = x$
$\implies \left(\frac{\sqrt{1 + {x}^{2}} - \sqrt{1 - {x}^{2}}}{\sqrt{1 + {x}^{2}} + \sqrt{1 - {x}^{2}}}\right) = \tan x$
$\implies \frac{\sqrt{1 + {x}^{2}} + \sqrt{1 - {x}^{2}}}{\sqrt{1 + {x}^{2}} - \sqrt{1 - {x}^{2}}} = \frac{1}{\tan} x$
$\implies \frac{\sqrt{1 + {x}^{2}}}{\sqrt{1 - {x}^{2}}} = \frac{1 + \tan x}{1 - \tan x}$ [ADDITION-DIVISION METHOD]
$\implies \frac{1 + {x}^{2}}{1 - {x}^{2}} = {\left(1 + \tan x\right)}^{2} / {\left(1 - \tan x\right)}^{2}$
=>1/x^2=((1+tanx)^2+(1-tanx)^2)/((1+tanx)^2-(1-tanx)^2
=>1/x^2=(2(tan^2x+1))/(4tanx#
$\implies {x}^{2} = \frac{2 \tan x}{{\tan}^{2} x + 1}$
$\implies {x}^{2} = 2 \tan x \cdot \frac{1}{\sec} ^ 2 x$
$\implies {x}^{2} = 2 \cdot \sin \frac{x}{\cos} x \cdot {\cos}^{2} x$
$\implies {x}^{2} = 2 \cdot \sin x \cdot \cos x$
$\implies x = \sqrt{\sin} 2 x$

Now what I have to solve...

Nov 14, 2017

$x = \sqrt{\sin 2 x}$

Explanation:

${\tan}^{-} 1 \left(\frac{\sqrt{1 + {x}^{2}} - \sqrt{1 - {x}^{2}}}{\sqrt{1 + {x}^{2}} + \sqrt{1 - {x}^{2}}}\right) = x$

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

$\tan x \cdot \left[\sqrt{1 + {x}^{2}} + \sqrt{1 - {x}^{2}}\right] = \sqrt{1 + {x}^{2}} - \sqrt{1 - {x}^{2}}$

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

$\left(1 + \tan x\right) \cdot \sqrt{1 - {x}^{2}} = \left(1 - \tan x\right) \cdot \sqrt{1 + {x}^{2}}$

${\left(1 + \tan x\right)}^{2} \cdot \left(1 - {x}^{2}\right) = {\left(1 - \tan x\right)}^{2} \cdot \left(1 + {x}^{2}\right)$

${\left(1 + \tan x\right)}^{2} - {x}^{2} \cdot {\left(1 + \tan x\right)}^{2} = {\left(1 - \tan x\right)}^{2} + {x}^{2} \cdot {\left(1 - \tan x\right)}^{2}$

${\left(1 + \tan x\right)}^{2} - {\left(1 - \tan x\right)}^{2} = {x}^{2} \cdot \left[{\left(1 + \tan x\right)}^{2} + {\left(1 - \tan x\right)}^{2}\right]$

${x}^{2} \cdot \left[2 {\left(\tan x\right)}^{2} + 2\right] = 4 \tan x$

$2 {x}^{2} \cdot \left[{\left(\tan x\right)}^{2} + 1\right] = 4 \tan x$

$2 {x}^{2} \cdot {\left(\sec x\right)}^{2} = 4 \tan x$

${x}^{2} = \frac{4 \tan x}{2 {\left(\sec x\right)}^{2}}$

${x}^{2} = 2 \tan x \cdot {\left(\cos x\right)}^{2}$

${x}^{2} = 2 \sin x \cdot \cos x$

${x}^{2} = \sin 2 x$

Hence $x = \sqrt{\sin 2 x}$