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, $x = tan ((alpha + beta)/2)$

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

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

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