Question

Prove that? : `arcsin ( (x-1)/(x+1) ) = 2arctan sqrt(x) - pi/2`

Answer 1

We wish to show that:

`arcsin ( (x-1)/(x+1) ) = 2arctan sqrt(x) - pi/2`

Define a function `f(x)` by:

`f(x) = arcsin ( (x-1)/(x+1) ) - 2arctan sqrt(x) + pi/2` ..... [A]

Put `x=0` and we get:

`f(0) = arcsin ( -1 ) - 2arctan 0 + pi/2`
`\ \ \ \ \ \ \ = -pi/2 - 0 + pi/2`
`\ \ \ \ \ \ \ = 0`

We can use the standard derivatives:

`{: (ul("Function"), ul("Derivative"), ul("Notes")), (f(x), f'(x),), (sin^(-1)x, 1/sqrt(1-x^2), ), (tan^(-1)x, 1/(1+x^2), ) :}`

So if we differentiate [A] and apply the chain rule, and quotient rule, we have:

`f'(x) = 1/sqrt(1-((x-1)/(x+1))^2)d/dx ((x-1)/(x+1)) - 2/(1+(sqrt(x))^2)d/dx(sqrt(x))`

`\ \ \ \ \ \ \ = 1/sqrt(1-(x-1)^2/(x+1)^2) ( (x+1)(1) - (x-1)(1) ) / (x+1)^2 - 2/(1+x) 1/(2sqrt(x))`

`\ \ \ \ \ \ \ = 1/sqrt( ( (x+1)^2 -(x-1)^2) /(x+1)^2) ( x+1 - x + 1 ) / (x+1)^2 - 1/((1+x)sqrt(x))`

`\ \ \ \ \ \ \ = sqrt( (x+1)^2 / ( (x+1)^2 -(x-1)^2) ) * ( 2 ) / (x+1)^2 - 1/((1+x)sqrt(x))`

`\ \ \ \ \ \ \ = sqrt( (x+1)^2 / ( (x+1+x-1)(x+1-x+1) ) ) * ( 2 ) / (x+1)^2 - 1/((1+x)sqrt(x))`

`\ \ \ \ \ \ \ = sqrt( (x+1)^2 / ( (2x)(2) ) ) * ( 2 ) / (x+1)^2 - 1/((1+x)sqrt(x))`

`\ \ \ \ \ \ \ = (x+1)/(2sqrt(x)) * ( 2 ) / (x+1)^2 - 1/((1+x)sqrt(x))`

`\ \ \ \ \ \ \ = 1/(sqrt(x)) * ( 1 ) / (x+1) - 1/((1+x)sqrt(x))`

`\ \ \ \ \ \ \ = 0`

So the function `f(x)` has a zero derivative everywhere, and therefore integrating we have:

`f(x) = c`

We established earlier that `f(0)=0 => f(0) = c = 0`, Hence:

`c = 0`

Hence we have:

`\ \ \ \ arcsin ( (x-1)/(x+1) ) - 2arctan sqrt(x) + pi/2 = 0`

`:. arcsin ( (x-1)/(x+1) ) = 2arctan sqrt(x) - pi/2` QED

Answer 2

Let `RHS=2tan^-1(sqrtx)-pi/2=y`

So `coty =cot(2tan^-1(sqrtx)-pi/2)`

`=>coty =-cot(pi/2-2tan^-1(sqrtx))`

`=>coty =-tan(2tan^-1(sqrtx))`

`=>coty =-(2tantan^-1(sqrtx))/(1-(tantan^-1(sqrtx) )^2`

`=>coty =-(2sqrt(x))/(1-((sqrtx) )^2`

`=>coty =-(2sqrt(x))/(1-x)`

`=>coty =(2sqrt(x))/(x-1)`

Now
`csc^2y=1+cot^2y`

`=>csc^2y=1+(4x)/(x-1)^2`

`=>csc^2y=((x-1)^2+4x)/(x-1)^2`

`=>csc^2y=(x+1)^2/(x-1)^2`

`=>cscy=(x+1)/(x-1)`

`=>siny=(x-1)/(x+1)`

`=>y=sin^-1((x-1)/(x+1))=LHS`