The range of operatorname{Sin}^{-1} x + opera | vedclass

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(C) We know that for $x \in [-1, 1]$,$\operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x = \frac{\pi}{2}$.Given expression is $f(x) = \operatorname

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$\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$

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(C) We know that for $x \in [-1, 1]$,$\operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x = \frac{\pi}{2}$.Given expression is $f(x) = \operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x + \operatorname{Tan}^{-1} x$.Substituting the identity,we get $f(x) = \frac{\pi}{2} + \operatorname{Tan}^{-1} x$.The domain of $\operatorname{Sin}^{-1} x$ and $\operatorname{Cos}^{-1} x$ is $[-1, 1]$.Therefore,the domain of $f(x)$ is $[-1, 1]$.We calculate the values at the boundaries:For $x = -1$,$f(-1) = \frac{\pi}{2} + \operatorname{Tan}^{-1}(-1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.For $x = 1$,$f(1) = \frac{\pi}{2} + \operatorname{Tan}^{-1}(1) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.Since $\operatorname{Tan}^{-1} x$ is a strictly increasing function,the range of $f(x)$ is $\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$.

The range of $\operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x + \operatorname{Tan}^{-1} x$ is

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