Inv Trig (1 Viewer)

Drongoski · Nov 24, 2022

Explanation. Please refer to your graphs of i) y = arctan(x) and ii) y = arccos(x)

$\text{For: }y = tan^{-1}x \text { x can be any real number. }\\ \\ \text {But in } y = tan^{-1}\sqrt{x^2 -1} $, $ \sqrt{x^2 -1} \geq 0 $ so that $ 0 \leq y <\frac {\pi}{2}\\ \\ \\$For: $ y = cos^{-1}(e^x) = cos^{-1}($ something $ > 0) $ the domain of arccos here is restricted to the set of all positive values only, so that the range is restricted to: $ 0 \leq y < \frac {\pi}{2}$

Inv Trig (1 Viewer)

Drongoski

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