Question #11b8e

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Question #11b8e

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Answer 1 · 2017-11-13T16:38:20

See the answer below...

Explanation:

${tan}^{- 1} (\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}) = x$ $tan^-1([sqrt{1+x^2}-sqrt{1-x^2}]/ [sqrt{1+x^2}+sqrt{1-x^2}])=x$
$\Rightarrow (\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}) = tan x$ $=>([sqrt{1+x^2}-sqrt{1-x^2}]/ [sqrt{1+x^2}+sqrt{1-x^2}])=tanx$
$\Rightarrow \frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}} = \frac{1}{tan x}$ $=>[sqrt{1+x^2}+sqrt{1-x^2}]/ [sqrt{1+x^2}-sqrt{1-x^2}]=1/tanx$
$\Rightarrow \frac{\sqrt{1 + x^{2}}}{\sqrt{1 - x^{2}}} = \frac{1 + tan x}{1 - tan x}$ $=>sqrt(1+x^2)/sqrt(1-x^2)=(1+tanx)/(1-tanx)$ [ADDITION-DIVISION METHOD]
$\Rightarrow \frac{1 + x^{2}}{1 - x^{2}} = \frac{{(1 + tan x)}^{2}}{{(1 - tan x)}^{2}}$ $=>(1+x^2)/(1-x^2)=(1+tanx)^2/(1-tanx)^2$
$\Rightarrow \frac{1}{x^{2}} = \frac{{(1 + tan x)}^{2} + {(1 - tan x)}^{2}}{{(1 + tan x)}^{2} - {(1 - tan x)}^{2}}$ $=>1/x^2=((1+tanx)^2+(1-tanx)^2)/((1+tanx)^2-(1-tanx)^2$
$\Rightarrow \frac{1}{x^{2}} = \frac{2 ({tan}^{2} x + 1)}{4 tan x}$ $=>1/x^2=(2(tan^2x+1))/(4tanx$
$\Rightarrow x^{2} = \frac{2 tan x}{{tan}^{2} x + 1}$ $=>x^2=(2tanx)/(tan^2x+1)$
$\Rightarrow x^{2} = 2 tan x \cdot \frac{1}{{sec}^{2} x}$ $=>x^2=2tanxcdot1/sec^2x$
$\Rightarrow x^{2} = 2 \cdot \frac{sin x}{cos x} \cdot {cos}^{2} x$ $=>x^2=2 cdot sinx/cosxcdotcos^2x$
$\Rightarrow x^{2} = 2 \cdot sin x \cdot cos x$ $=>x^2=2 cdot sinx cdot cosx$
$\Rightarrow x = \sqrt{sin 2} x$ $=>x=sqrtsin2x$

Now what I have to solve...

Answer 2 · 2017-11-14T19:19:35

$x = \sqrt{sin 2 x}$ $x=sqrt(sin2x)$

Explanation:

${tan}^{- 1} (\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}) = x$ $tan^-1([sqrt{1+x^2}-sqrt{1-x^2}]/ [sqrt{1+x^2}+sqrt{1-x^2}])=x$

$tan x = \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}$ $tanx=[sqrt{1+x^2}-sqrt{1-x^2}]/ [sqrt{1+x^2}+sqrt{1-x^2}]$

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

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

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

${(1 + tan x)}^{2} \cdot (1 - x^{2}) = {(1 - tan x)}^{2} \cdot (1 + x^{2})$ $(1+tanx)^2*(1-x^2)=(1-tanx)^2*(1+x^2)$

${(1 + tan x)}^{2} - x^{2} \cdot {(1 + tan x)}^{2} = {(1 - tan x)}^{2} + x^{2} \cdot {(1 - tan x)}^{2}$ $(1+tanx)^2-x^2*(1+tanx)^2=(1-tanx)^2+x^2*(1-tanx)^2$

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

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

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

$2 x^{2} \cdot {(sec x)}^{2} = 4 tan x$ $2x^2*(secx)^2=4tanx$

$x^{2} = \frac{4 tan x}{2 {(sec x)}^{2}}$ $x^2=(4tanx)/[2(secx)^2]$

$x^{2} = 2 tan x \cdot {(cos x)}^{2}$ $x^2=2tanx*(cosx)^2$

$x^{2} = 2 sin x \cdot cos x$ $x^2=2sinx*cosx$

$x^{2} = sin 2 x$ $x^2=sin2x$

Hence $x = \sqrt{sin 2 x}$ $x=sqrt(sin2x)$

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