Answer
$f(0)f(1)<0,f$ continuous $\Rightarrow \exists x_0\in (0,1),f(x_0)=0$
Work Step by Step
We are given the function:
$f(x)=\tan^{-1} x-\cos^{-1} x$
First we determine the domain of $f(x)$.
The domain of $\tan^{-1} x$ is $(-\infty,\infty)$, while the domain of $\cos^{-1} x$ is $[-1,1]$,therefore the domain of $f$ is:
$(-\infty,\infty)\cap [-1,1]=[-1,1]$
Choose two points in the domain and compute the value of the function in them:
$f(0)=\tan^{-1} 0-\cos^{-1} 0=-\dfrac{\pi}{2}$
$f(1)=\tan^{-1} 1-\cos^{-1} 1=\dfrac{\pi}{4}$
As $f(x)$ is continuous on $(-1,1)$ and $f(0)$ and $f(1)$ have opposite signs, it means that the function has a zero between 0 and 1.
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