Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

The domain of ${\sec }^{-1}(x)$ and ${\mathrm{cosec}}^{-1}(x)$ is $(-\infty ,-1]\cup [1,\infty )$. The domain of ${\tan }^{-1}(x)$ is $(-\infty ,\infty )$. Therefore, the domain of $f(x)$ is $(-\infty ,-1]\cup [1,\infty )$. We know the property ${\sec }^{-1}(x)+{\mathrm{cosec}}^{-1}(x)=\frac{\pi }{2}$ for $x\in (-\infty ,-1]\cup [1,\infty )$. So, $f(x)=\frac{\pi }{2}+{\tan }^{-1}(x)$. Case 1: If $x\in [1,\infty )$, then ${\tan }^{-1}(x)\in [{\tan }^{-1}(1),{\lim }_{x\to \infty }{\tan }^{-1}(x))=[\frac{\pi }{4},\frac{\pi }{2})$. Thus, $f(x)\in [\frac{\pi }{2}+\frac{\pi }{4},\frac{\pi }{2}+\frac{\pi }{2})=[\frac{3\pi }{4},\pi )$. Case 2: If $x\in (-\infty ,-1]$, then ${\tan }^{-1}(x)\in ({\lim }_{x\to -\infty }{\tan }^{-1}(x),{\tan }^{-1}(-1)]=(-\frac{\pi }{2},-\frac{\pi }{4}]$. Thus, $f(x)\in (\frac{\pi }{2}-\frac{\pi }{2},\frac{\pi }{2}-\frac{\pi }{4}]=(0,\frac{\pi }{4}]$. Combining both cases, the range of $f(x)$ is $(0,\frac{\pi }{4}]\cup [\frac{3\pi }{4},\pi )$.