If `sin^(-1)((2a)/(1+a^2))+sin^(-1)((2b)/(1+b^2))=2tan^(-1)x ,` then `x` is equal to `[a , b , in (0,1)]` `(a-b)/(1+a b)` (b) `b/(1+a b)` (c) `b/(1+a b)` (d) `(a+b)/(1-a b)`

Options

A. `(a - b)/(1 + ab)`
B. `(b)/(1 + ab)`
C. `(b)/(1 - ab)`
D. `(a + b)/(1 - ab)`

Answer

D

Explanation

Since `sin^(-1) ((2x)/(1 + x^(2))) = 2 tan^(-1) x " for " x in (-1, 1)`
`sin^(-1) ((2a)/(1 + a^(2))) + sin^(-1) ((2b)/(1 + b^(2))) = 2 tan^(-1) x`
`rArr 2 tan^(-1) a + 2 tan^(-1) b = 2 tan^(-1) x`
or `tan^(-1) a + 2 tan^(-1) b = 2 tan^(-1) x`
or `tan^(-1) ((a + b)/(1 -ab)) = tan^(-1) x`
or `x = (a + b)/(1 - ab)`