Step-by-step Solution:

Concept:

Double angle formula:

\( \sin 2x = \dfrac{2 \tan x}{1 + \tan^2 x} \)

Addition formula:

\( \tan(x + y) = \dfrac{\tan x + \tan y}{1 - \tan x \tan y} \)

 

Calculation:

The given identity is: \( \sin^{-1}\left( \dfrac{2a}{1 + a^2} \right) + \sin^{-1}\left( \dfrac{2b}{1 + b^2} \right) = 2 \tan^{-1} n \).

Let \( a = \tan y_1 \) and \( b = \tan y_2 \).

Therefore, the given equation becomes:

\( \sin^{-1}\left( \dfrac{2(\tan y_1)}{1 + (\tan y_1)^2} \right) + \sin^{-1}\left( \dfrac{2(\tan y_2)}{1 + (\tan y_2)^2} \right) = 2 \tan^{-1} n \).

\( \sin^{-1}(\sin 2y_1) + \sin^{-1}(\sin 2y_2) = 2 \tan^{-1} n \).

\( 2y_1 + 2y_2 = 2 \tan^{-1} n \).

\( y_1 + y_2 = \tan^{-1} n \).

\( \tan(y_1 + y_2) = n \).

\( \dfrac{\tan y_1 + \tan y_2}{1 - \tan y_1 \tan y_2} = n \).

\( \dfrac{a + b}{1 - ab} = n \).

Therefore, \( n = \dfrac{a + b}{1 - ab} \).