How do you solve #arcsin[(2a)/(1+a^2)] + arcsin[(2b)/(1+b^2)] = 2arctanx#?

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Answer 1 · 2016-06-20T12:36:09

# x = ( a + b )/( 1 - a b )#

Use #sin theta = (2 tan (theta/2))/(1+tan^2(theta/2)# and

#tan ( u + v )=( tan u + tan v )/( 1 - tan u tan v )#.

Let #a = tan (alpha/2) and b = tan (beta/2)#.

Then, the given equation becomes

#alpha + beta = 2 (arc tan x)#.

So, #x = tan ((alpha + beta)/2)#

#=(tan (alpha/2)+tan(beta/2))/( 1 - tan (alpha/2) tan (beta/2)#

#=(a+b)/(1-ab)# .