If \(\rm \sin^{-1} \dfrac{2a}{1+a^2} + \sin^{-1} \dfrac{2b}{1+b^2}=2

1.	If \(\rm \sin^{-1} \dfrac{2a}{1+a^2} + \sin^{-1} \dfrac{2b}{1+b^2}=2 \tan^{-1} n\) then?1. \(n=\dfrac{a-b}{1+ab}\)2. \(n=\dfrac{(ab)}{(a-a)}\)3. \(n=\dfrac{(a+b)}{(1-ab)}\)4. \(n=\dfrac{(1-ab)}{(1+ab)}\)
Answer» 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, \(\rm n = \frac{a+b}{1-ab}\).

If \(\rm \sin^{-1} \dfrac{2a}{1+a^2} + \sin^{-1} \dfrac{2b}{1+b^2}=2 \tan^{-1} n\) then?1. \(n=\dfrac{a-b}{1+ab}\)2. \(n=\dfrac{(ab)}{(a-a)}\)3. \(n=\dfrac{(a+b)}{(1-ab)}\)4. \(n=\dfrac{(1-ab)}{(1+ab)}\)

Answer» 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, \(\rm n = \frac{a+b}{1-ab}\).