2009/027) If α = 2 arctan [(1 + x)/(1 - x)] and β = arcsin [(1 - x^2)/(1 + x^2)], then what is α + β

we know tan(pi/4+y) = tan (pi/4 + tan y)/(1 tan pi/4-tan y)

so tan (pi/4+y) = (1+ tan y)/(1-tan y))

so pi/4 + y = tan ^-1(1+tan y)/(1-tan y)

put x = tan y

so pi/4 + y = arctan ((1+x)/(1-x))

α = 2 arctan ((1+x)/(1-x)) = pi/2 + 2 tan ^- 1 x .1

again cos 2y = cos ^2 y - sin ^2 y

= cos^2 y(1- tan ^2 y)
= (1- tan ^2 y)/sec^2 y
= (1- tan ^2 y)/(1+ tan ^2y)

so put tan y = x

cos 2y = (1-x^2)/(1+x^2)
2y = arccos ((1-x^2)/(1+x^2))

so arcsin ((1-x^2)/(1+x^2)) = pi/2 - 2y = pi/2 - 2 tan ^- x .2

addimg 1 and 2 we get
α + β = π