Question #11b8e

2 Answers
Nov 13, 2017

See the answer below...

Explanation:

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

Now what I have to solve...

Nov 14, 2017

x=sqrt(sin2x)

Explanation:

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

tanx=[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}

tanx*sqrt(1+x^2)+tanx*sqrt(1-x^2)=sqrt{1+x^2}-sqrt{1-x^2}

(1+tanx)*sqrt(1-x^2)=(1-tanx)*sqrt(1+x^2)

(1+tanx)^2*(1-x^2)=(1-tanx)^2*(1+x^2)

(1+tanx)^2-x^2*(1+tanx)^2=(1-tanx)^2+x^2*(1-tanx)^2

(1+tanx)^2-(1-tanx)^2=x^2*[(1+tanx)^2+(1-tanx)^2]

x^2*[2(tanx)^2+2]=4tanx

2x^2*[(tanx)^2+1]=4tanx

2x^2*(secx)^2=4tanx

x^2=(4tanx)/[2(secx)^2]

x^2=2tanx*(cosx)^2

x^2=2sinx*cosx

x^2=sin2x

Hence x=sqrt(sin2x)