Answer:
[tex]y=cos^{-1}(x)[/tex],
[tex]y=cot^{-1}(-\frac{1}{2})[/tex],
[tex]y=sin^{-1}(-\frac{1}{2})[/tex]
[tex]y=tan^{-1}(-\frac{1}{2})[/tex]
Above four functions are defined at the given point.
Step-by-step explanation:
We have been given all trigonometric function we need to tell which among them is defined for [tex]x=-\frac{1}{2}[/tex]
Case 1: [tex]y=cos^{-1}(x)[/tex]
Since, [tex]At x=-\frac{1}{2}[/tex]
[tex]y=cos^{-1}(-\frac{1}{2})[/tex]
[tex]y=-cos^{-1}(\frac{1}{2})[/tex]
[tex]y=\frac{2\pi}{3}[/tex]
Domain of the function is [tex][-1,1][/tex]
Defined at given point.
Case 2: [tex]y=cot^{-1}(-\frac{1}{2})[/tex]
[tex]y=-cot^{-1}(\frac{1}{2})[/tex]
[tex]y=-1.1071[/tex]
Domain of the function is [tex](-\infty,\infty)[/tex]
Defined at given point.
Case 3: [tex]y=cosec^{-1}(-\frac{1}{2})[/tex]
[tex]y=-cosec^{-1}(\frac{1}{2})[/tex]
Value of the function is not defined because it is out of the domain of the function.
Domain is [tex](-\infty,-1]\cup[1,\infty)[/tex]
So, not defined at given point.
Case 4: [tex]y=sec^{-1}(-\frac{1}{2})[/tex]
[tex]y=-sec^{-1}(\frac{1}{2})[/tex]
Value of the function is not defined because it is out of the domain of the function.
Domain is [tex](-\infty,-1]\cup[1,\infty)[/tex]
So, not defined at given point.
Case 5:[tex]y=sin^{-1}(-\frac{1}{2})[/tex]
[tex]y=-sin^{-1}(\frac{1}{2})[/tex]
[tex]y=\frac{-\pi}{6}[/tex]
Domain of the function is [tex][-1,1][/tex]
Defined at given point.
Case 6:[tex]y=tan^{-1}(-\frac{1}{2})[/tex]
[tex]y=-tan^{-1}(\frac{1}{2})[/tex]
[tex]y=-0.4636[/tex]
Domain of the function is [tex](-\infty,\infty)[/tex]
Defined at given point.
Therefore, Option 1,2,5,6 are correct.
