Question

Question #11b8e

Answer 1

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...

Answer 2

`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)`