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Find range of the function:$f(x)=(\sin^{-1}x)^2-(\cot^{-1}x)^2$

The domain of the function is $-1\leq x \leq 1$

$f(-1)=(\sin^{-1}(-1))^2-(\cot^{-1}(-1))^2=\frac{\pi^2}{4}-\frac{9\pi^2}{16}=-\frac{-5\pi^2}{16}$

$f(1)=\frac{\pi^2}{4}-\frac{\pi^2}{16}=\frac{3\pi^2}{16}$

But I am not able to show that $f(x)$ is monotonic.

$f'(x)=\frac{2\sin^{-1}x}{\sqrt{1-x^2}}+\frac{2\cot^{-1}x}{1+x^2}$

How to go from here.

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    $\begingroup$ Using a change of variable $\theta=\sin^{-1}x$ may simplify the expression a bit? $\endgroup$
    – Feng
    Dec 7, 2023 at 4:50
  • $\begingroup$ Why do you think $f$ is monotonic? Just looking at it, it seems unlikely to be to me. $\endgroup$ Dec 8, 2023 at 1:02

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