range of inverse trig. function

Question

Find the range of $f(x)$

$$f(x) = \sec^{-1}x + \csc^{-1}x + \cot^{-1}x$$

$\Rightarrow$ $$f(x)= \sec^{-1}x + \left({\pi\over 2} - \sec^{-1}x\right) + \cot^{-1}x $$

$\Rightarrow$ $$f(x)= {\pi\over 2} + \cot^{-1}x$$

Answer 1

The domain of the given function $f(x)$ is $x\in(-\infty,-1)\cup(1,\infty)$. Hence, $$\cot^{-1}{x}\in \left(0,\dfrac{\pi}{4}\right)\cup\left(\dfrac{3\pi}{4},\pi\right)$$

Now, as per your simplified expression for $f(x)$ which is $f(x)={\pi\over 2} + \cot^{-1}x$.

The range comes out to be $$f(x)\in\left(\dfrac{\pi}{2},\dfrac{3\pi}{4}\right)\cup\left(\dfrac{5\pi}{4},\dfrac{3\pi}{2}\right)$$

Answer 2

According to your final

⇒$f(x)={\pi\over 2}+cot^{−1}x$

range of $cot^{-1}x = (0,π)$ and if ${\pi\over 2}$ is added the range become $\left({\pi\over 2},{3\pi\over2}\right)$