# 3.14 Inverse trigonometric functions  (Page 3/4)

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$f:R-\left(-1,1\right)\to \left[-\frac{\pi }{2},\frac{\pi }{2}\right]-\left\{0\right\}\phantom{\rule{1em}{0ex}}\text{by f(x)}=\text{arccosec (x)}$

The arccosec(x) .vs. x graph is shown here.

## Arcsecant function

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.

$\text{Domain of secant}=\left[0,\pi /2\right)\cup \left(\pi /2,\pi \right]=\left[0,\pi \right]-\left\{\pi /2\right\}$

$\text{Range of secant}=\left(-\infty ,-1\right]\cup \left[1,\infty \right)=R-\left(-1,1\right)$

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 :

$\text{Domain of arcsecant}=R-\left(-1,1\right)$

$\text{Range of arcsecant}=\left[0,\pi \right]-\left\{\pi /2\right\}$

Therefore, we define arcsecant function as :

$f:R-\left(-1,1\right)\to \left[0,\pi \right]-\left\{\pi /2\right\}\phantom{\rule{1em}{0ex}}\text{by f(x)}=\text{arcsec (x)}$

The arcsec(x) .vs. x graph is shown here.

## Arccotangent function

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.

$\text{Domain of cotangent}=\left(0,\pi \right)$

$\text{Range of cotangent}=R$

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 :

$\text{Domain of arccotangent}=R$

$\text{Range of arccotangent}=\left(0,\pi \right)$

Therefore, we define arccotangent function as :

$f:R\to \left(0,\pi \right)\phantom{\rule{1em}{0ex}}\text{by f(x)}=\text{arccot (x)}$

The arccot(x) .vs. x graph is shown here.

## Example

Problem : Find y when :

$y=\mathrm{tan}{}^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

Solution : There are multiple angles for which :

$⇒\mathrm{tan}y=x=-\frac{1}{\sqrt{3}}$

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 :

$⇒y=2\pi -\frac{\pi }{6}=\frac{11\pi }{6}$

Corresponding negative angle is :

$⇒y=\frac{11\pi }{6}-2\pi =-\frac{\pi }{6}$

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 :

$D={D}_{1}\cap {D}_{2}-\left\{x:x\phantom{\rule{1em}{0ex}}\text{when h(x)}=0\right\}$

Here, $g\left(x\right)={\mathrm{cos}}^{-1}\left(x\right)$ . The domain of arccosine is [-1,1]. Hence,

${D}_{1}=\text{Domain of “g”}=\left[-1,1\right]$

The denominator function h(x) is greatest integer function. Its domain is “R”.

${D}_{2}=\text{Domain of “h”}=R$

The intersection of two domains is :

$⇒{D}_{1}\cap {D}_{2}=\left[-1,1\right]\cap R=\left[-1,1\right]$

Now, greatest integer function becomes zero for values of “x” in the interval [0,1). Hence, domain of given function is :

$D={D}_{1}\cap {D}_{2}-\left[0,1\right)$

$D=\left[-1,1\right]-\left[0,1\right)=-1\le x<0\cup \left\{1\right\}$

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