If -1 ≤ x ≤ 0, then $tan \begin{Bmatrix}\frac{1}{2}sin^{-1}\frac{2x}{1+x^2}+\frac{1}{2}cos^{-1}\frac{1-x^2}{1+x^2}\end{Bmatrix}$ is equal to
Answer & explanation
Correct answer: option 2
We know that
$sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2 tan^{-1}x$, if $-1≤ x≤ 1$
and,
$cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = -2tan^{-1}x,$ if $-∞ < x ≤ 0.$
$∴ tan \begin{Bmatrix}\frac{1}{2}sin^{-1}\frac{2x}{1+x^2}+\frac{1}{2}cos^{-1}\frac{1-x^2}{1+x^2}\end{Bmatrix}$
$= tan (tan^{-1} x - tan^{-1} x), $ if $-1 ≤ x ≤ 0 = tan 0=0$