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