Answer
sec$^{-1}(2)\approx1.05$ radians
csc$^{-1}(3)\approx0.34$ radians
cot$^{-1}(4)\approx0.25$ radians
Work Step by Step
To prove sec$^{-1}(x)=$cos$^{-1}(\frac{1}{x})$, we are going to assign $y$ to cos$^{-1}(\frac{1}{x})$ So now we'll solve for $y$:
sec$^{-1}(x)=y$
$x=$sec$(y)$
$x=\frac{1}{cos(y)}$
cos$(y)=\frac{1}{x}$
$y=$cos$^{-1}(\frac{1}{x})$
And, indeed, we see that it has been proven. The others have a very similar procedure.
To prove csc$^{-1}(x)=$sin$^{-1}(\frac{1}{x})$, we are going to assign $y$ to sin$^{-1}(\frac{1}{x})$ So now we'll solve for $y$:
csc$^{-1}(x)=y$
$x=$csc$(y)$
$x=\frac{1}{sin(y)}$
sin$(y)=\frac{1}{x}$
$y=$sin$^{-1}(\frac{1}{x})$
And, indeed, we see that it has been proven.
To prove cot$^{-1}(x)=$tan$^{-1}(\frac{1}{x})$, we are going to assign $y$ to tan$^{-1}(\frac{1}{x})$ So now we'll solve for $y$:
cot$^{-1}(x)=y$
$x=$cot$(y)$
$x=\frac{1}{tan(y)}$
tan$(y)=\frac{1}{x}$
$y=$tan$^{-1}(\frac{1}{x})$
And, indeed, we see that it has been proven.
