# Inv Trig (1 Viewer)

#### Drongoski

##### Well-Known Member
I'm late but how did u find the domain and range for the second question? And the range for the first
But I still don't know if my answers were correct?

#### Pethmin

##### Active Member
But I still don't know if my answers were correct?
Sorry forgot to confirm, yes they (all answers you provided) are correct

#### Pethmin

##### Active Member
I'm late but how did u find the domain and range for the second question? And the range for the first
Range for 1st: using composite function again (let u=sqrtx^2-1) therefore, y=arctanu- find common restriction for 'u' and use both graphs to using the restriction to determine range and Domain

#### Drongoski

##### Well-Known Member
Explanation. Please refer to your graphs of i) y = arctan(x) and ii) y = arccos(x)

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

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