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$$f:R-\left(-\mathrm{1,}1\right)\to [-\frac{\pi}{2},\frac{\pi}{2}]-\left\{0\right\}\phantom{\rule{1em}{0ex}}\text{by f(x)}=\text{arccosec (x)}$$
The arccosec(x) .vs. x graph is shown here.
The arcsecant function is inverse function of trigonometric secant function. From the plot of secant function, it is clear that union of two disjointed intervals between “0 and $\pi /2$ ” and “ $\pi /2$ and $\pi $ ” includes all possible values of secant function only once. Note that “ $\pi /2$ ” is excluded. The redefinition of domain of trigonometric function, however, does not change the range.
$$\text{Domain of secant}=[\mathrm{0,}\pi /2)\cup (\pi /\mathrm{2,}\pi ]=\left[\mathrm{0,}\pi \right]-\{\pi /2\}$$
$$\text{Range of secant}=(-\infty ,-1]\cup \left[\mathrm{1,}\infty \right)=R-\left(-\mathrm{1,1}\right)$$
This redefinition renders secant function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :
$$\text{Domain of arcsecant}=R-\left(-\mathrm{1,}1\right)$$
$$\text{Range of arcsecant}=\left[\mathrm{0,}\pi \right]-\{\pi /2\}$$
Therefore, we define arcsecant function as :
$$f:R-\left(-\mathrm{1,}1\right)\to \left[\mathrm{0,}\pi \right]-\{\pi /2\}\phantom{\rule{1em}{0ex}}\text{by f(x)}=\text{arcsec (x)}$$
The arcsec(x) .vs. x graph is shown here.
The arccotangent function is inverse function of trigonometric cotangent function. From the plot of cotangent function it is clear that an interval between 0 and $\pi $ includes all possible values of cotangent function only once. Note that end points are excluded. The redefinition of domain of trigonometric function, however, does not change the range.
$$\text{Domain of cotangent}=\left(\mathrm{0,}\pi \right)$$
$$\text{Range of cotangent}=R$$
This redefinition renders cotangent function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :
$$\text{Domain of arccotangent}=R$$
$$\text{Range of arccotangent}=\left(\mathrm{0,}\pi \right)$$
Therefore, we define arccotangent function as :
$$f:R\to \left(\mathrm{0,}\pi \right)\phantom{\rule{1em}{0ex}}\text{by f(x)}=\text{arccot (x)}$$
The arccot(x) .vs. x graph is shown here.
Problem : Find y when :
$$y=\mathrm{tan}{}^{-1}\left(-\frac{1}{\sqrt{3}}\right)$$
Solution : There are multiple angles for which :
$$\Rightarrow \mathrm{tan}y=x=-\frac{1}{\sqrt{3}}$$
However, range of sine function is [-π/2, π/2]. We need to find angle, which falls in this range. Now, acute angle corresponding to the value of 1/√3 is π/6. In accordance with sign diagram, tangent is negative in second and fourth quarters. But range is [-π/2, π/2]. Hence, we need to find angle in fourth quadrant. The angle in the fourth quadrant whose tangent has magnitude of 1/√3 is given by :
$$\Rightarrow y=2\pi -\frac{\pi}{6}=\frac{11\pi}{6}$$
Corresponding negative angle is :
$$\Rightarrow y=\frac{11\pi}{6}-2\pi =-\frac{\pi}{6}$$
Problem : Find domain of the function given by :
$$f\left(x\right)=\frac{{\mathrm{cos}}^{-1}\left(x\right)}{\left[x\right]}$$
Solution : The given function is quotient of two functions having rational form :
$$f\left(x\right)=\frac{g\left(x\right)}{h\left(x\right)}$$
The domain of quotient is given by :
$$D={D}_{1}\cap {D}_{2}-\{x:x\phantom{\rule{1em}{0ex}}\text{when h(x)}=0\}$$
Here, $g\left(x\right)={\mathrm{cos}}^{-1}\left(x\right)$ . The domain of arccosine is [-1,1]. Hence,
$${D}_{1}=\text{Domain of \u201cg\u201d}=[-\mathrm{1,1}]$$
The denominator function h(x) is greatest integer function. Its domain is “R”.
$${D}_{2}=\text{Domain of \u201ch\u201d}=R$$
The intersection of two domains is :
$$\Rightarrow {D}_{1}\cap {D}_{2}=[-\mathrm{1,1}]\cap R=[-\mathrm{1,1}]$$
Now, greatest integer function becomes zero for values of “x” in the interval [0,1). Hence, domain of given function is :
$$D={D}_{1}\cap {D}_{2}-\left[\mathrm{0,1}\right)$$
$$D=[-\mathrm{1,1}]-\left[\mathrm{0,1}\right)=-1\le x<0\cup \left\{1\right\}$$
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