2arctan(sqrtx)-arcsin((x-1)/(x+1)) is equal to pi/2?

it is defined for x >= 0

now let y = arctan(sqrtx)

cos 2arctan(sqrtx)
= cos (2y)
= (cos^2 y - sin ^2 y)/(cos^2 y + sin ^2 y)
= (cos^2 y - sin ^2 y)/(cos^2 y + sin ^2 y) as denominator is 1
= ( 1- tan ^2 y)/(1+ tan ^2 y)
= (1-tan ^2 y)/(1+tan ^2 y)
= (1-x)/(x+1)

so 2arctan(sqrtx)-arcsin((x-1)/(x+1)
= arc cos((1-x)/(x+1)) - arcsin((x-1)/(x+1))
= arc cos((1-x)/(x+1)) + arcsin((1-x)/(x+1)) as arcsin t = - arcsin - t
= pi/2 as arccos t + arcsin t = pi/2

(if t >= 0 both in 1st quadrant and sum is between 0 and pi)
if ( t < 0 then arc cos is between pi/2 and pi and arc sin < 0) so sum between 0 and pi) this is mentioned for fixing the range