Correct Answer - Option 3 :
\(n=\dfrac{(a+b)}{(1-ab)}\)
Concept:
Double angle formula:
\(\rm \sin2x=\dfrac{2\tan x}{1+\tan^2}\)
Addition formula:
\(\rm \tan(x+y) = \dfrac{\tan x + \tan y}{1-\tan x\tan y}\)
Calculation:
The given identity is \(\rm \sin^{-1}\left(\frac{2a}{1+a^2}\right) + \sin^{-1}\left(\frac{2b}{1+b^2}\right) = 2\tan^{-1}n\).
Let \(\rm a = \tan y_1\mbox{ and } b= \tan y_2\).
Therefore, the given equation becomes:
\(\rm \sin^{-1}\left(\frac{2a}{1+a^2}\right) + \sin^{-1}\left(\frac{2b}{1+b^2}\right)= 2\tan^{-1}n\)
\(\rm \sin^{-1}\left(\frac{2(\tan y_1)}{1+(\tan y_1)^2}\right) + \sin^{-1}\left(\frac{2(\tan y_2)}{1+(\tan y_2)^2}\right) = 2\tan^{-1}n\)
\(\rm \sin^{-1}\left(\sin 2y_1\right)+\sin^{-1}\left(\sin 2y_2\right) = 2\tan^{-1}n\)
\(\rm 2y_1 +2y_2 = 2\tan^{-1}n\)
\( \rm y_1 +y_2 = \tan^{-1}n \)
\(\rm \tan(y_1 + y_2) = n\)
\(\rm \frac{\tan y_1 + \tan y_2}{1-\tan y_1\tan y_2}=n\)
\(\rm \frac{a+b}{1-ab}=n \)
Therefore, \(\rm n = \frac{a+b}{1-ab}\).
