The range of

Question:

The range of $\sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x$ is _______________________.

Solution:

Domain of the given function $=[-1,1] \cap \mathbf{R}=[-1,1]$

Now,

For $-1 \leq x \leq 1$

$\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ and $-\frac{\pi}{4} \leq \tan ^{-1} x \leq \frac{\pi}{4}$

$\therefore \frac{\pi}{2}-\frac{\pi}{4} \leq \sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x \leq \frac{\pi}{2}+\frac{\pi}{4}$

$\Rightarrow \frac{\pi}{4} \leq \sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x \leq \frac{3 \pi}{4}$

Thus, the range of the given function is $\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]$.

The range of $\sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x$ is $\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]$