Let y = ^ -1 ((1-x^2)/(1+x^2)) - ^ -1 ((2x)/(1+x^2)) . The set of...

We know the principal value branches of the given inverse trigonometric functions:

$\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right) =$\begin{cases} 2\tan^{-1}x, & x \ge 0 \\ -2\tan^{-1}x, & x < 0 \end{cases}<br><br>$\sin^{-1}\left(\dfrac{2x}{1+x^2}\right) = $\begin{cases} -\pi - 2\tan^{-1}x, & x < -1 \\ 2\tan^{-1}x, & -1 \le x \le 1 \\ \pi - 2\tan^{-1}x, & x > 1 \end{cases}

Now, we evaluate $y = \cos^{-1}$\left(\dfrac{1-x^2}{1+x^2}\right) - \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$in different intervals:<br><br>Case 1:$x \in (-\infty, -1)$<br>$y = (-2\tan^{-1}x) - (-\pi - 2\tan^{-1}x) = \pi$<br>Since$y$is a constant,$\dfrac{dy}{dx} = 0$.<br><br>Case 2:$x \in (-1, 0)$<br>$y = (-2\tan^{-1}x) - (2\tan^{-1}x) = -4\tan^{-1}x$<br>Here,$\dfrac{dy}{dx} = \dfrac{-4}{1+x^2} \ne 0$.<br><br>Case 3:$x \in (0, 1)$<br>$y = (2\tan^{-1}x) - (2\tan^{-1}x) = 0$<br>Since$y$is a constant,$\dfrac{dy}{dx} = 0$.<br><br>Case 4:$x \in (1, \infty)$<br>$y = (2\tan^{-1}x) - (\pi - 2\tan^{-1}x) = 4\tan^{-1}x - \pi$<br>Here,$\dfrac{dy}{dx} = \dfrac{4}{1+x^2} \ne 0$.<br><br>Thus,$\dfrac{dy}{dx} = 0$for$x \in (-\infty, -1) \cup (0, 1)$.<br><br>Answer:$(-\infty, -1) \cup (0, 1)