Respuesta :
Answer: C) [tex]y=sin^-1 x[/tex]
Step-by-step explanation:
Since, the graph of an inverse trigonometric function will pass through the point [tex](1,\frac{\pi}{2})[/tex],
If this point satisfies the function,
For the function [tex]y=cos^{-1} x[/tex]
If x = 1
[tex]y=cos^{-1}1=0[/tex]
Thus, [tex](1,\frac{\pi}{2})[/tex] is not satisfying function [tex]y=cos^{-1} x[/tex],
⇒ The graph of [tex]y=cos^{-1} x[/tex] is not passing through the point [tex](1,\frac{\pi}{2})[/tex]
For the function [tex]y=cot^{-1}x[/tex]
If x = 1
[tex]y=cot^{-1}1=\frac{\pi}{4}[/tex]
Thus, [tex](1,\frac{\pi}{2})[/tex] is not satisfying function [tex]y=cot^{-1}x[/tex],
⇒ The graph of [tex]y=cot^{-1}x[/tex] is not passing through the point [tex](1,\frac{\pi}{2})[/tex]
For the function [tex]y=sin^{-1} x[/tex]
If x = 1
[tex]y=sin^{-1}1=\frac{\pi}{2}[/tex]
Thus, [tex](1,\frac{\pi}{2})[/tex] is satisfying function [tex]y=sin^{-1} x[/tex],
⇒ The graph of [tex]y=sin^{-1} x[/tex] is passing through the point [tex](1,\frac{\pi}{2})[/tex].
For the function [tex]y=tan^{-1}x[/tex]
If x = 1
[tex]y=tan^{-1}1=\frac{\pi}{4}[/tex]
Thus, [tex](1,\frac{\pi}{2})[/tex] is not satisfying function [tex]y=cos^{-1} x[/tex],
⇒ The graph of [tex]y=tan^{-1} x[/tex] is not passing through the point [tex](1,\frac{\pi}{2})[/tex].
Hence, Option C is correct.
