18.6 - Solved Examples on Inverse Trigonometric Functions

In the previous sections, we have seen all the six inverse trigonometric functions. The following table will help us to memorize the principal branch of each of those functions.

$\begin{array}{cc}{}    &{\textbf{Function}}    &{\textbf{Domain}}    &{\textbf{Range}}    &{}\\
{}    &{\sin^{-1}}    &{[-1,1]}    &{\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]}    &{}\\
{}    &{\cos^{-1}}    &{[-1,1]}    &{\left[0, \pi \right]}    &{}\\
{}    &{\csc^{-1}}    &{R - (-1,1)}    &{\left[\frac{- \pi}{2}, \frac{\pi}{2} \right] - \{0 \}}    &{}\\
{}    &{\sec^{-1}}    &{R - (-1,1)}    &{\left[0, \pi \right] - \{\frac{\pi}{2} \}}    &{}\\
{}    &{\tan^{-1}}    &{R}    &{\left(\frac{- \pi}{2}, \frac{\pi}{2} \right)}    &{}\\
{}    &{\cot^{-1}}    &{R}    &{\left(0, \pi \right)}    &{}\\
\end{array}$

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Let us write three important points to remember:
1. We denote the inverse trigonometric functions using the superscript '-1'.
• For example, the inverse sine function is denoted as sin^-1.
• This should not be confused with (sin x)^-1.
(sin x)^-1 is $\frac{1}{\sin x}$
• This is applicable to all trigonometric functions.
2. If the branch is not specified, it is understood that, the principal branch is being considered.
3. Consider any one of the six inverse trigonometric functions.
• That function will have only one set as it’s domain.
• But that function will have infinite number of sets as the range.
• If we pick a value from the domain and use it as the input, we will get an output in each of the range sets.
• But the output present in the range corresponding to the principal branch, is considered as the principal value of that function.

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Now we will see some solved examples   
Solved Example 18.1
Find the principal value of $\sin^{-1} \left(\frac{1}{\sqrt{2}} \right)$
Solution:
1. We are asked to find $\sin^{-1} \left(\frac{1}{\sqrt{2}} \right)$
• Let $x~=~\sin^{-1} \left(\frac{1}{\sqrt{2}} \right)$
2. So our aim is to find x. It can be done in 4 steps:
(i) $x~=~\sin^{-1} \left(\frac{1}{\sqrt{2}} \right)$ is an equation of the form:
$x~=~f(y)~=~\sin^{-1}(y)$
(ii) This is an inverse trigonometric function, where input y = $\frac{1}{\sqrt2}$.
• Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~ \sin x$
• In our present case, it is: $y ~=~ \frac{1}{\sqrt2} ~=~ \sin x$
(iii) So we have a trigonometric equation:
$\sin x = \frac{1}{\sqrt2}$
• When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11.
• In the present case, we do not need to write many steps. We already know that, $\sin \left(\frac{\pi}{4} \right)~=~\frac{1}{\sqrt2}$
• So we can write: $x~=~\frac{\pi}{4}$
3. Finally, we check whether the value obtained is the principal value. It can be done in 5 steps:
(i) The final answer that we obtained is: $\sin^{-1} \left(\frac{1}{\sqrt{2}} \right)~=~\frac{\pi}{4}$
(ii) It is clear that,
• For the given inverse trigonometric function,
   ♦ The input y is $\frac{1}{\sqrt2}$
   ♦ The output x is $\frac{\pi}{4}$
(iii) For the $\sin^{-1}$ function:
   ♦ Domain is [-1,1]
   ♦ Range corresponding to the principal branch is $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$
(iv) The input y falls within [-1,1]. So the input is acceptable.
(v) The output x falls within $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$. So the output obtained is also acceptable.
• We can write:
   ♦ The output $\frac{\pi}{4}$,
   ♦ is the principal value of the $\sin^{-1}$ function,
   ♦ when the input is $\frac{1}{\sqrt2}$.

Solved Example 18.2
Find the principal value of $\cot^{-1} \left(\frac{-1}{\sqrt{3}} \right)$
Solution:
1. We are asked to find $\cot^{-1} \left(\frac{-1}{\sqrt{3}} \right)$
• Let $x~=~\cot^{-1} \left(\frac{-1}{\sqrt{3}} \right)$
2. So our aim is to find x. It can be done in 4 steps:
(i) $x~=~\cot^{-1} \left(\frac{-1}{\sqrt{3}} \right)$ is an equation of the form:
$x~=~f(y)~=~\cot^{-1}(y)$
(ii) This is an inverse trigonometric function, where input y = $\frac{-1}{\sqrt3}$.
• Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~ \cot x$
• In our present case, it is: $y ~=~ \frac{-1}{\sqrt3} ~=~ \cot x$
(iii) So we have a trigonometric equation:
$\cot x = \frac{-1}{\sqrt3}$
• When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11. In the present case, it can be done as shown below:

$\begin{array}{ll}{}    &{\cot x}    & {~=~}    &{\frac{-1}{\sqrt{3}}}    &{} \\
{\implies}    &{\tan x}    & {~=~}    &{-\sqrt{3}~~\color{magenta}{\text{- - - (A)}}}    &{} \\
{}    &{\tan \left(\frac{\pi}{3} \right)}    & {~=~}    &{\sqrt{3}~~\color{magenta}{\text{- - - (B)}}}    &{} \\
{}    &{\left[\tan \left(\pi – \theta \right)\right.}    & {~=~}    &{\left. - \tan \theta\right] ~~\color{magenta}{\text{- - - (C)}}}    &{} \\
{\implies}    &{\tan \left(\pi – \frac{\pi}{3} \right)}    & {~=~}    &{- \tan \frac{\pi}{3}}    &{} \\
{\implies}    &{\tan \left(\frac{2 \pi}{3} \right)}    & {~=~}    &{- \sqrt{3}~~\color{magenta}{\text{- - - (D)}}}    &{} \\
{\implies}    &{x}    & {~=~}    &{\frac{2 \pi}{3}~~\color{magenta}{\text{- - - (E)}}}    &{} \\
\end{array}               
$

◼ Remarks:
• Line A:
For simplicity, we convert cot x to tan x. This line gives the modified equation which is to be solved.
• Line B:
We write a basic equation which is closest to the equation to be solved.
• Line C:
We write the trigonometric identity which will help to solve the equation.
This identity is derived from identities 9(c) and 9(d). The list of identities can be seen here.
• Line D:
We get this result from (B).
• Line E:
We get this result by comparing D and A.

3. Finally, we check whether the value obtained is the principal value. It can be done in 5 steps:
(i) The final answer that we obtained is: $\cot^{-1} \left(\frac{-1}{\sqrt{3}} \right)~=~\frac{2 \pi}{3}$
(ii) Based on this, we can write:
• For the given inverse trigonometric function,
   ♦ The input y is $\frac{-1}{\sqrt3}$
   ♦ The output x is $\frac{2 \pi}{3}$
(iii) For the $\cot^{-1}$ function:
   ♦ Domain is R
   ♦ Range corresponding to the principal branch is $\left(0, \pi \right)$
(iv) The input y is a real number. So the input is acceptable.
(v) The output x falls within $\left(0, \pi \right)$. So the output obtained is also acceptable.
• We can write:
   ♦ The output $\frac{2 \pi}{3}$,
   ♦ is the principal value of the $\cot^{-1}$ function,
   ♦ when the input is $\frac{-1}{\sqrt3}$.

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The link below gives a few more solved examples:

Exercise 18.1

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In the next section, we will see properties of inverse trigonometric functions.

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18.5 - Inverse of Cotangent Function

In the previous section, we saw tan^-1 function. In this section, we will see cot^-1 function.

Some basics can be written in 8 steps:
1. Consider the cot function:
f(x) = cot x
◼ For this function, we must choose the input values carefully. It can be written in 3 steps:
(i) We know that: $\cot x = \frac{\cos x}{\sin x}$.
• The denominator should not become zero. That means, sin x should not become zero.

(ii) We know that:
    ♦ sin 0 = 0
    ♦ sin π = 0
    ♦ sin (-π) = 0
    ♦ sin 2π = 0
    ♦ sin (-2π) = 0
    ♦ sin 3π = 0
    ♦ so on . . .
• So we can write:
The input x should not be equal to nπ, where n is an integer.

(iii) Thus we get the domain for f(x) = cot x as:
set {x : x∈R and x ≠ nπ, n∈Z}
• That means, x can be any real number except nπ, where n is an integer.

◼ Similarly, we must have a good knowledge about the codomain of cot x. It can be written in 3 steps:
(i) We know that, output of sin x will lie in the interval [-1,1].
• That means, output of sin x will be any one of the four items below:
    ♦ -1
    ♦ a -ve proper fraction
    ♦ a -ve proper fraction
    ♦ +1
(Though zero lies in the interval [-1,1], we are not allowing sin x to become zero. We achieve this by avoiding nπ as input x values)

(ii) Based on the above "possible outputs of sin x", we can write the "possible outputs of cot x":
• Remember that, the numerator cos x can give any output in the interval [-1,1].
So the output of $\frac{\cos x}{\sin x}$ can be any real number.
For example:
$\cot x = \frac{\cos x}{\sin x} = \frac{\frac{7}{67}}{\frac{98}{99}} ~=~ 0.1051$

(iii) Thus we get the codomain of cot x as: R

• We saw the above details in class 11. We saw a neat pictorial representation of the above details in the graph of the cot function. It is shown again in fig.18.22 below:

Fig.18.22

2. Let us check whether the cot function is one-one.
• Let input x = $\frac{\pi}{2}$.
    ♦ Then the output will be $f \left(\frac{\pi}{2} \right)~=~\cot \left(\frac{\pi}{2} \right)~=~0 $
• Let input x = $\frac{- \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{- \pi}{2} \right)~=~\cot \left(\frac{- \pi}{2} \right)~=~0 $
• Let input x = $\frac{-3 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{-3 \pi}{2} \right)~=~\cot \left(\frac{-3 \pi}{2} \right)~=~0 $  

(We can cross check with the graph and confirm that the above inputs and outputs are correct)

• We see that, more than one input values from the domain can give the same output. So the cot function is not a one-one function.

3. Suppose that, we restrict the input values.
• That is, we take input values only from the set $\left(0, \pi \right)$.
    ♦ Note that, we use '()' instead of '[]'.
    ♦ That means, the boundary values should not be used as inputs.

• We already saw the codomain. It is: R
• Then we will get the green curve shown in fig.18.23 below:

Fig.18.23

• In the green portion, no two inputs will give the same output. So the green portion represents a function which is one-one.

4. Next, we have to prove that, the green portion is onto.
• For that, we can consider any y value from the codomain R.
• There will be always a x value in $\left(0, \pi \right)$, which will satisfy the equation y = cot x.
• So the green portion is onto.
5. We see that, the green portion is both one-one and onto.
• We can represent this function in the mathematical way:
$\text{cot}:~ \left(0, \pi \right)~\to~R$, defined as f(x) = cot x.
6. If a function is both one-one and onto, the codomain is same as range.
• So we can write:
For this function,
    ♦ the domain is $\left(0, \pi \right)$
    ♦ the range is R
7. We know that, if a function is one-one and onto, it will be invertible. We have seen the properties of inverse functions. Let us apply those properties to our present case. It can be written in 4 steps:
(i) If y = f(x) = cot x is invertible, then there exists a function g such that: g(y) = x
(ii) The function g will also be one-one and onto.
(iii) The domain of f will be the range of g. So the range of g is $\left(0, \pi \right)$.
(iv) The range of f will be the domain of g. So the domain of g is R
(iv) The inverse of cot function is denoted as cot^-1. So we can define the inverse function as:
$\cot^{-1}:~ R~\to~\left(0, \pi \right)$, defined as x = g(y) = cot^-1 y.
8. Let us see an example:
• Suppose that, for the inverse function, the input y is $\sqrt{3}$
• Then we get an equation: $x ~=~\cot^{-1} \left(\sqrt{3} \right)$
• Our aim is to find x. It can be done in 4 steps:
(i) $x ~=~\cot^{-1} \left(\sqrt{3} \right)$ is an equation of the form $x ~=~f(y)~=~\cot^{-1} \left(y \right)$
(ii) This is an inverse trigonometric function, where input y = $\sqrt{3}$.
• Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~\cot x$
• In our present case, it is:
$y~=~\sqrt{3}~=~\cot x$
(iii) So we have a trigonometric equation:
$\cot x~=~\sqrt{3}$
• When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11.
• In the present case, we do not need to write many steps. We already know that, $\cot \left(\frac{\pi}{6} \right)~=~\sqrt{3}$
• So we can write:
$x ~=~\frac{\pi}{6}$

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The above 8 steps help us to understand the basics about cosec^-1 function. Now we will see a few more details. It can be written in 4 steps:
1. We saw that, the cot function is not a one-one function. But to make it one-one, we restricted the domain to $\left(0, \pi \right)$.
2. There are other possible “restricted domains” available.
• $\left(- \pi, 0 \right)$ is shown in magenta color in fig.18.24 below:

Fig.18.24

• $\left(-2 \pi, -\pi \right)$ is shown in cyan color in fig,18.12 above.
3. There are infinite number of such restricted domains possible.
• We say that:
    ♦ Each restricted domain gives a corresponding branch of the cot^-1 function.
    ♦ The restricted domain $\left(0, \pi \right)$ gives the principal branch of the cot^-1 function.

4. In class 11, we plotted the cot function. Now we will plot the inverse. It can be done in 4 steps:
(i) Write the set for the original function f. It must contain a convenient number of ordered pairs.
• $\left(\frac{\pi}{6} , \sqrt{3} \right)$ is an example of the ordered pairs in f.
(To get a smooth curve, we must write a large number of ordered pairs)
(ii) Based on set f, we can write set g.
This is done by picking each ordered pair from f and interchanging the positions.
• For example, the point $\left(\frac{\pi}{6} , \sqrt{3}  \right)$ in the set f will become $\left(\sqrt{3} , \frac{\pi}{6}  \right)$ in set g.
• Thus we will get the required number of ordered pairs in g.
(iii) Mark each ordered pair of g on the graph paper.
• The first coordinate should be marked along the x-axis.
• The second coordinate should be marked along the y-axis.
(iv) Once all the ordered pairs are marked, draw a smooth curve connecting all the marks.
• The smooth curve is the required graph. It is shown in fig.18.25 below:

Fig.18.25

(v) Unlike the graphs of sin^-1 and cos^-1, the graph of cosec^-1 is not smaller in width. This is because:
Any real number, can be used as input values. That means, the graph can extend upto -∞ towards the left and upto +∞ towards the right.
• The graph is larger in height because:
    ♦ Depending upon the branch, values upto +∞ or –∞ can be obtained as output values.
• The green curve is related to the $\left(0, \pi \right)$ branch.
• The cyan curve is related to the $\left(\pi, 2 \pi \right)$ branch. 
• The magenta curve is related to the $\left(- \pi, 0 \right)$ branch.

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We have seen all the six inverse trigonometric functions. In the next section, we will see some solved examples.

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18.4 - Inverse of Tangent Function

In the previous section, we saw sec^-1 function. In this section, we will see tan^-1 function.

Some basics can be written in 8 steps:
1. Consider the tan function:
f(x) = tan x
◼ For this function, we must choose the input values carefully. It can be written in 3 steps:
(i) We know that: $\tan x = \frac{\sin x}{\cos x}$.
• The denominator should not become zero. That means, cos x should not become zero.

(ii) We know that:
    ♦ $\cos \left(\frac{\pi}{2} \right)$ = 0
    ♦ $\cos \left(\frac{3 \pi}{2} \right)$ = 0
    ♦ $\cos \left(\frac{- \pi}{2} \right)$ = 0
    ♦ $\cos \left(\frac{-3 \pi}{2} \right)$ = 0
    ♦ $\cos \left(\frac{5 \pi}{2} \right)$ = 0
    ♦ so on . . .
• So we can write:
The input x should not be equal to $(2n+1) \frac{\pi}{2}$, where n is an integer.

(iii) Thus we get the domain for f(x) = sec x as:
set $\{x:x \in R ~\text{and}~x \ne (2n+1)\frac{\pi}{2},~n \in Z \}$
• That means, x can be any real number except $(2n+1) \frac{\pi}{2}$, where n is an integer.

◼ Similarly, we must have a good knowledge about the codomain of the tan function. It can be written in 3 steps:
(i) We know that, output of cos x will lie in the interval [-1,1].
• That means, output of cos x will be any one of the four items below:
    ♦ -1
    ♦ a -ve proper fraction
    ♦ a -ve proper fraction
    ♦ +1
(Though zero lies in the interval [-1,1], we are not allowing cos x to become zero. We achieve this by avoiding $(2n+1) \frac{\pi}{2}$ as input x values)

(ii) Based on the above possible outputs of cos x, we can write the possible outputs of tan x:
• Remember that, the numerator sin x can give any output in the interval [-1,1].
• So the output of $\frac{\sin x}{\cos x}$ can be any real number.
For example:
$\tan x = \frac{\sin x}{\cos x} = \frac{\frac{98}{99}}{\frac{7}{67}} ~=~ 9.514$

(iii) Thus we get the codomain of tan x as: R

• We saw the above details in class 11. We saw a neat pictorial representation of the above details in the graph of the tan function. It is shown again in fig.18.18 below:

Fig.18.18

2. Let us check whether the tan function is one-one.
• Let input x = $\frac{\pi}{4}$
    ♦ Then the output will be $f \left( \frac{\pi}{4} \right)~=~\tan \left( \frac{\pi}{4} \right)~=~1 $
• Let input x = $ \frac{-3 \pi}{4} $.
    ♦ Then the output will be $f \left( \frac{-3 \pi}{4} \right)~=~\tan \left( \frac{-3 \pi}{4} \right)~=~1 $
• Let input x = $ \frac{5 \pi}{4} $  
    ♦ Then the output will be $f \left( \frac{5 \pi}{4} \right)~=~\tan \left( \frac{5 \pi}{4} \right)~=~1 $

(We can cross check with the graph and confirm that the above inputs and outputs are correct)

• We see that, more than one input values from the domain can give the same output. So the tan function is not a one-one function.

3. Suppose that, we restrict the input values.
• That is, we take input values only from the set $\left(\frac{- \pi}{2}, \frac{\pi}{2} \right)$.
    ♦ Note that, we use '()' instead of '[]'.
    ♦ That means, the boundary values should not be used as inputs.
• We already saw the codomain. It is: R
• Then we will get the green curves shown in fig.18.19 below:

Fig.18.19

• In the green portion, no two inputs will give the same output. So the green portion represents a function which is one-one.

4. Next, we have to prove that, the green portion is onto.
• For that, we can consider any y value from the codomain R.
• There will be always a x value in $\left(\frac{- \pi}{2}, \frac{\pi}{2} \right)$, which will satisfy the equation y = tan x.
• So the green portion is onto.
5. We see that, the green portion is both one-one and onto.
• We can represent this function in the mathematical way:
$\text{tan}:~ \left(\frac{- \pi}{2}, \frac{\pi}{2} \right)~\to~R$, defined as f(x) = tan x.
6. If a function is both one-one and onto, the codomain is same as range.
• So we can write:
For this function, the domain is $\left(\frac{- \pi}{2}, \frac{\pi}{2} \right)$ and range is R.
7. We know that, if a function is one-one and onto, it will be invertible. We have seen the properties of inverse functions. Let us apply those properties to our present case. It can be written in 4 steps:
(i) If y = f(x) = tan x is invertible, then there exists a function g such that:
g(y) = x
(ii) The function g will also be one-one and onto.
(iii) The domain of f will be the range of g. So the range of g is $\left(\frac{- \pi}{2}, \frac{\pi}{2} \right)$.
(iv) The range of f will be the domain of g. So the domain of g is R.
(iv) The inverse of tan function is denoted as tan^-1. So we can define the inverse function as:
$\tan^{-1}:~ R ~\to~\left(\frac{- \pi}{2}, \frac{\pi}{2} \right)$, defined as x = g(y) = tan^-1 y.
8. Let us see an example:
• Suppose that, for the inverse function, the input y is $\frac{1}{\sqrt{3}}$
• Then we get an equation: $x ~=~\tan^{-1} \left( \frac{1}{\sqrt{3}} \right)$
• Our aim is to find x. It can be done in 4 steps:
(i) $x ~=~\tan^{-1} \left(\frac{1}{\sqrt{3}} \right)$ is an equation of the form $x ~=~f(y)~=~\tan^{-1} \left(y \right)$
(ii) This is an inverse trigonometric function, where input y = $\frac{1}{\sqrt{3}}$.
• Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~\tan x$
• In our present case, it is:
$y~=~\frac{1}{\sqrt{3}}~=~\tan x$
(iii) So we have a trigonometric equation:
$\tan x~=~\frac{1}{\sqrt{3}}$
• When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11.
• In the present case, we do not need to write many steps. We already know that, $\tan \left(\frac{\pi}{6} \right)~=~\frac{1}{\sqrt{3}}$
• So we can write:
$x ~=~\frac{\pi}{6}$

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The above 8 steps help us to understand the basics about tan^-1 function. Now we will see a few more details. It can be written in 4 steps:
1. We saw that, the tan function is not a one-one function. But to make it one-one, we restricted the domain to $\left(\frac{- \pi}{2}, \frac{\pi}{2} \right)$.
2. There are other possible “restricted domains” available.
• $\left(\frac{-3 \pi}{2}, \frac{- \pi}{2} \right)$ is shown in magenta color in fig.18.20 below:

Fig.18.20

• $\left(\frac{ \pi}{2}, \frac{3 \pi}{2} \right)$ is shown in cyan color in fig,18.20 above.
3. There are infinite number of such restricted domains possible.
• We say that:
    ♦ Each restricted domain gives a corresponding branch of the tan^-1 function.
    ♦ The restricted domain $\left(\frac{- \pi}{2}, \frac{\pi}{2} \right)$ gives the principal branch of the tan^-1 function.

4. In class 11, we plotted the tangent function. Now we will plot the inverse. It can be done in 4 steps:
(i) Write the set for the original function f. It must contain a convenient number of ordered pairs.
• $\left(\frac{\pi}{3} , \sqrt{3}  \right)$ is an example of the ordered pairs in f.
(To get a smooth curve, we must write a large number of ordered pairs)
(ii) Based on set f, we can write set g.
This is done by picking each ordered pair from f and interchanging the positions.
• For example, the point $\left(\frac{\pi}{3} , \sqrt{3}  \right)$ in the set f will become $\left(\sqrt{3} , \frac{\pi}{3}  \right)$ in set g.
• Thus we will get the required number of ordered pairs in g.
(iii) Mark each ordered pair of g on the graph paper.
• The first coordinate should be marked along the x-axis.
• The second coordinate should be marked along the y-axis.
(iv) Once all the ordered pairs are marked, draw a smooth curve connecting all the marks.
• The smooth curve is the required graph. It is shown in fig.18.21 below:

Fig.18.21

(v) Unlike the graphs of sin^-1 and cos^-1, the graph of tan^-1 is not smaller in width. This is because:
Any real number, can be used as input values. That means, the graph can extend upto -∞ towards the left and upto +∞ towards the right.
• The graph is larger in height because:
    ♦ Depending upon the branch, values upto +∞ or –∞ can be obtained as output values.
• The green curve is related to the $\left(\frac{- \pi}{2}, \frac{\pi}{2} \right)$ branch.
• The cyan curve is related to the $\left(\frac{ \pi}{2}, \frac{3 \pi}{2} \right)$ branch. 
• The magenta curve is related to the $\left(\frac{-3 \pi}{2}, \frac{- \pi}{2} \right)$ branch.

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In the next section, we will see cot^-1 function.

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18.3 - Inverse of sec Function

In the previous section, we saw cosec^-1 function. In this section, we will see sec^-1 function.

Some basics can be written in 8 steps:
1. Consider the sec function:
f(x) = sec x
◼ For this function, we must choose the input values carefully. It can be written in 3 steps:
(i) We know that: $\sec x = \frac{1}{\cos x}$.
• The denominator should not become zero. That means, cos x should not become zero.

(ii) We know that:
    ♦ $\cos \left(\frac{\pi}{2} \right)$ = 0
    ♦ $\cos \left(\frac{3 \pi}{2} \right)$ = 0
    ♦ $\cos \left(\frac{- \pi}{2} \right)$ = 0
    ♦ $\cos \left(\frac{-3 \pi}{2} \right)$ = 0
    ♦ $\cos \left(\frac{5 \pi}{2} \right)$ = 0
    ♦ so on . . .
• So we can write:
The input x should not be equal to $(2n+1) \frac{\pi}{2}$, where n is an integer.

(iii) Thus we get the domain for f(x) = sec x as:
set $\{x:x \in R ~\text{and}~x \ne (2n+1)\frac{\pi}{2},~n \in Z \}$
• That means, x can be any real number except $(2n+1) \frac{\pi}{2}$, where n is an integer.

◼ Similarly, we must have a good knowledge about the codomain. It can be written in 3 steps:
(i) We know that, output of cos x will lie in the interval [-1,1].
• That means, output of cos x will be any one of the four items below:
    ♦ -1
    ♦ a -ve proper fraction
    ♦ a -ve proper fraction
    ♦ +1
(Though zero lies in the interval [-1,1], we are not allowing sin x to become zero. We achieve this by avoiding $(2n+1) \frac{\pi}{2}$ as input x values)

(ii) Based on the above possible outputs of cos x, we can write the possible outputs of sec x:
    ♦ If it is -1, the output will be -1.
    ♦ If it is a -ve proper fraction, the output will be real number smaller than -1.
    ♦ If it is a +ve proper fraction, the output will be real number larger than 1.
    ♦ If it is +1, the output will be +1.

• We see that, sec x will not give outputs in the interval (-1,1). Note that, for writing this interval, we use '()' instead of '[]'. This is because, -1 and +1 are not included in the interval. Those two values can become outputs.

(iii) Thus we get the codomain:
R − (−1,1)

• We saw the above details in class 11. We saw a neat pictorial representation of the above details in the graph of the sec function. It is shown again in fig.18.14 below:

Fig.18.14

2. Let us check whether the sec function is one-one.
• Let input x = 0
    ♦ Then the output will be $f \left( 0 \right)~=~\sec \left( 0 \right)~=~1 $
• Let input x = $ \pi $.
    ♦ Then the output will be $f \left(2 \pi \right)~=~\sec \left(2 \pi \right)~=~1 $   
• Let input x = $-2 \pi $.
    ♦ Then the output will be $f \left(-2 \pi \right)~=~\sec \left(-2 \pi \right)~=~1 $

(We can cross check with the graph and confirm that the above inputs and outputs are correct)

• We see that, more than one input values from the domain can give the same output. So the sec function is not a one-one function.

3. Suppose that, we restrict the input values.
• That is, we take input values only from the set $\left[0, \pi \right]$.
    ♦ But $\frac{\pi}{2}$ lies in this interval. It cannot be an input.
    ♦ So the modified set is: $\left[0, \pi \right]~-~\{ \frac{\pi}{2} \}$
• We already saw the codomain. It is: R − (−1,1)
• Then we will get the green curves shown in fig.18.15 below:

Fig.18.15

• In the green portion, no two inputs will give the same output. So the green portion represents a function which is one-one.

4. Next, we have to prove that, the green portion is onto.
• For that, we can consider any y value from the codomain R − (−1,1).
• There will be always a x value in $\left[0, \pi \right]~-~\{ \frac{\pi}{2} \}$, which will satisfy the equation y = sec x.
• So the green portion is onto.
5. We see that, the green portion is both one-one and onto.
• We can represent this function in the mathematical way:
$\text{sec}:~ \left[0, \pi \right]~-~\{ \frac{\pi}{2} \}~\to~R~-~(-1,1)$, defined as f(x) = sec x.
6. If a function is both one-one and onto, the codomain is same as range.
• So we can write:
For this function, the domain is $\left[0, \pi \right]~-~\{ \frac{\pi}{2} \}$ and range is R − (−1,1)
7. We know that, if a function is one-one and onto, it will be invertible. We have seen the properties of inverse functions. Let us apply those properties to our present case. It can be written in 4 steps:
(i) If y = f(x) = sec x is invertible, then there exists a function g such that:
g(y) = x
(ii) The function g will also be one-one and onto.
(iii) The domain of f will be the range of g. So the range of g is $\left[0, \pi \right]~-~\{ \frac{\pi}{2} \}$.
(iv) The range of f will be the domain of g. So the domain of g is R − (−1,1)
(iv) The inverse of sec function is denoted as sec^-1. So we can define the inverse function as:
$\sec^{-1}:~ R − (−1,1)~\to~\left[0, \pi \right]~-~\{ \frac{\pi}{2} \}$, defined as x = g(y) = sec^-1 y.
8. Let us see an example:
• Suppose that, for the inverse function, the input y is 2
• Then we get an equation: $x ~=~\sec^{-1} \left(2 \right)$
• Our aim is to find x. It can be done in 5 steps:
(i) $x ~=~\sec^{-1} \left(2 \right)$ is an equation of the form $x ~=~f(y)~=~\sec^{-1} \left(y \right)$
(ii) This is an inverse trigonometric function, where input y = 2.
• Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~\sec x$
• In our present case, it is:
$y~=~2~=~\sec x$
(iii) So we have a trigonometric equation:
$\sec x~=~2$
• When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11.
• In the present case, we do not need to write many steps. We already know that, $\sec \left(\frac{\pi}{3} \right)~=~2$
• So we can write:
$x ~=~\frac{\pi}{3}$

---

The above 8 steps help us to understand the basics about sec^-1 function. Now we will see a few more details. It can be written in 4 steps:
1. We saw that, the sec function is not a one-one function. But to make it one-one, we restricted the domain to $\left[0, \pi \right]~-~\{ \frac{\pi}{2} \}$.
2. There are other possible “restricted domains” available.
• $\left[- \pi, 0 \right]~-~\{ \frac{- \pi}{2} \}$ is shown in magenta color in fig.18.16 below:

Fig.18.16

• $\left[\pi, 2 \pi \right]~-~\{ \frac{3 \pi}{2} \}$ is shown in cyan color in fig.18.16 above.
3. There are infinite number of such restricted domains possible.
• We say that:
    ♦ Each restricted domain gives a corresponding branch of the sec^-1 function.
    ♦ The restricted domain $\left[0, \pi \right]~-~\{ \frac{\pi}{2} \}$ gives the principal branch of the sec^-1 function.

4. In class 11, we plotted the sec function. Now we will plot the inverse. It can be done in 4 steps:
(i) Write the set for the original function f. It must contain a convenient number of ordered pairs.
• $\left(\frac{\pi}{6} , 2 \right)$ is an example of the ordered pairs in f.
(To get a smooth curve, we must write a large number of ordered pairs)
(ii) Based on set f, we can write set g.
This is done by picking each ordered pair from f and interchanging the positions.
• For example, the point $\left(\frac{\pi}{6} , 2  \right)$ in the set f will become $\left(2 , \frac{\pi}{6}  \right)$ in set g.
• Thus we will get the required number of ordered pairs in g.
(iii) Mark each ordered pair of g on the graph paper.
• The first coordinate should be marked along the x-axis.
• The second coordinate should be marked along the y-axis.
(iv) Once all the ordered pairs are marked, draw a smooth curve connecting all the marks.
• The smooth curve is the required graph. It is shown in fig.18.17 below:

Fig.18.17

(v) Unlike the graphs of sin^-1 and cos^-1, the graph of sec^-1 is not smaller in width. This is because:
Any real number except those in the interval (-1,1), can be used as input values. That means, the graph can extend upto -∞ towards the left and upto +∞ towards the right.
• The graph is larger in height because:
    ♦ Depending upon the branch, values upto +∞ or –∞ can be obtained as output values.
• The green curve is related to the $\left[0, \pi \right]~-~\{ \frac{\pi}{2} \}$ branch.
• The cyan curve is related to the $\left[\pi, 2 \pi \right]~-~\{ \frac{3 \pi}{2} \}$ branch. 
• The magenta curve is related to the $\left[- \pi, 0 \right]~-~\{ \frac{- \pi}{2} \}$ branch.

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In the next section, we will see tan^-1 function.

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18.2 - Inverse of cosec Function

In the previous section, we saw cos^-1 function. In this section, we will see cosec^-1 function.

Some basics can be written in 8 steps:
1. Consider the cosec function:
f(x) = cosec x
◼ For this function, we must choose the input values carefully. It can be written in 3 steps:
(i) We know that: $\csc x = \frac{1}{\sin x}$.
• The denominator should not become zero. That means, sin x should not become zero.

(ii) We know that:
    ♦ sin 0 = 0
    ♦ sin π = 0
    ♦ sin (-π) = 0
    ♦ sin 2π = 0
    ♦ sin (-2π) = 0
    ♦ sin 3π = 0
    ♦ so on . . .
• So we can write:
The input x should not be equal to nπ, where n is an integer.

(iii) Thus we get the domain for f(x) = cosec x as:
set {x : x∈R and x ≠ nπ, n∈Z}
• That means, x can be any real number except nπ, where n is an integer.

◼ Similarly, we must have a good knowledge about the codomain. It can be written in 3 steps:
(i) We know that, output of sin x will lie in the interval [-1,1].
• That means, output of sin x will be any one of the four items below:
    ♦ -1
    ♦ a -ve proper fraction
    ♦ a -ve proper fraction
    ♦ +1
(Though zero lies in the interval [-1,1], we are not allowing sin x to become zero. We achieve this by avoiding nπ as input x values)

(ii) Based on the above "possible outputs of sin x", we can write the "possible outputs of cosec x":
    ♦ If it is -1, the output will be -1.
    ♦ If it is a -ve proper fraction, the output will be real number smaller than -1.
    ♦ If it is a +ve proper fraction, the output will be real number larger than 1.
    ♦ If it is +1, the output will be +1.

• We see that, cosec x will not give outputs in the interval (-1,1). Note that, for writing this interval, we use '()' instead of '[]'. This is because, -1 and +1 are not included in the interval. Those two values can become outputs.

(iii) Thus we get the codomain:
R − (−1,1)

• We saw the above details in class 11. We saw a neat pictorial representation of the above details in the graph of the cosec function. It is shown again in fig.18.10 below:

Fig.18.10

2. Let us check whether the cosec function is one-one.
• Let input x = $\frac{\pi}{2}$.
    ♦ Then the output will be $f \left(\frac{\pi}{2} \right)~=~\csc \left(\frac{\pi}{2} \right)~=~1 $
• Let input x = $\frac{5 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{5 \pi}{2} \right)~=~\csc \left(\frac{5 \pi}{2} \right)~=~1 $   
• Let input x = $\frac{-3 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{-3 \pi}{2} \right)~=~\csc \left(\frac{-3 \pi}{2} \right)~=~1 $

(We can cross check with the graph and confirm that the above inputs and outputs are correct)

• We see that, more than one input values from the domain can give the same output. So the cosec function is not a one-one function.

3. Suppose that, we restrict the input values.
• That is, we take input values only from the set $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]$.
    ♦ But zero lies in this interval. Zero cannot be an input.
    ♦ So the modified set is: $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]~-~\{ 0 \}$
• We already saw the codomain. It is: R − (−1,1)
• Then we will get the green curves shown in fig.18.11 below:

Fig.18.11

• In the green portion, no two inputs will give the same output. So the green portion represents a function which is one-one.

4. Next, we have to prove that, the green portion is onto.
• For that, we can consider any y value from the codomain R − (−1,1).
• There will be always a x value in $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]~-~\{ 0 \}$, which will satisfy the equation y = cosec x.
• So the green portion is onto.
5. We see that, the green portion is both one-one and onto.
• We can represent this function in the mathematical way:
$\text{cosec}:~ \left[\frac{- \pi}{2}, \frac{\pi}{2} \right]~-~\{ 0 \}~\to~R~-~(-1,1)$, defined as f(x) = cosec x.
6. If a function is both one-one and onto, the codomain is same as range.
• So we can write:
For this function,
    ♦ the domain is $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right] - \{ 0 \}$
    ♦ the range is R − (−1,1)
7. We know that, if a function is one-one and onto, it will be invertible. We have seen the properties of inverse functions. Let us apply those properties to our present case. It can be written in 4 steps:
(i) If y = f(x) = cosec x is invertible, then there exists a function g such that: g(y) = x
(ii) The function g will also be one-one and onto.
(iii) The domain of f will be the range of g. So the range of g is $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]~-~\{ 0 \}$.
(iv) The range of f will be the domain of g. So the domain of g is R − (−1,1)
(iv) The inverse of cosec function is denoted as cosec^-1. So we can define the inverse function as:
$\csc^{-1}:~ R − (−1,1)~\to~\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]~-~\{0 \}$, defined as x = g(y) = cosec^-1 y.
8. Let us see an example:
• Suppose that, for the inverse function, the input y is 2
• Then we get an equation: $x ~=~\csc^{-1} \left(2 \right)$
• Our aim is to find x. It can be done in 4 steps:
(i) $x ~=~\csc^{-1} \left(2 \right)$ is an equation of the form $x ~=~f(y)~=~\csc^{-1} \left(y \right)$
(ii) This is an inverse trigonometric function, where input y = 2.
• Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~\csc x$
• In our present case, it is:
$y~=~2~=~\csc x$
(iii) So we have a trigonometric equation:
$\csc x~=~2$
• When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11.
• In the present case, we do not need to write many steps. We already know that, $\csc \left(\frac{\pi}{6} \right)~=~2$
• So we can write:
$x ~=~\frac{\pi}{6}$

---

The above 8 steps help us to understand the basics about cosec^-1 function. Now we will see a few more details. It can be written in 4 steps:
1. We saw that, the cosec function is not a one-one function. But to make it one-one, we restricted the domain to $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right] - \{ 0 \}$.
2. There are other possible “restricted domains” available.
• $\left[\frac{- 3 \pi}{2}, \frac{- \pi}{2} \right] - \{- \pi \}$ is shown in magenta color in fig.18.12 below:

Fig.18.12

• $\left[\frac{\pi}{2}, \frac{3 \pi}{2} \right] - \{\pi \}$ is shown in cyan color in fig,18.12 above.
3. There are infinite number of such restricted domains possible.
• We say that:
    ♦ Each restricted domain gives a corresponding branch of the cosec^-1 function.
    ♦ The restricted domain $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right] - \{ 0 \}$ gives the principal branch of the cosec^-1 function.

4. In class 11, we plotted the cosine function. Now we will plot the inverse. It can be done in 4 steps:
(i) Write the set for the original function f. It must contain a convenient number of ordered pairs.
• $\left(\frac{\pi}{6} , 2 \right)$ is an example of the ordered pairs in f.
(To get a smooth curve, we must write a large number of ordered pairs)
(ii) Based on set f, we can write set g.
This is done by picking each ordered pair from f and interchanging the positions.
• For example, the point $\left(\frac{\pi}{6} , 2  \right)$ in the set f will become $\left(2 , \frac{\pi}{6}  \right)$ in set g.
• Thus we will get the required number of ordered pairs in g.
(iii) Mark each ordered pair of g on the graph paper.
• The first coordinate should be marked along the x-axis.
• The second coordinate should be marked along the y-axis.
(iv) Once all the ordered pairs are marked, draw a smooth curve connecting all the marks.
• The smooth curve is the required graph. It is shown in fig.18.13 below:

Fig.18.13

(v) Unlike the graphs of sin^-1 and cos^-1, the graph of cosec^-1 is not smaller in width. This is because:
Any real number except those in the interval (-1,1), can be used as input values. That means, the graph can extend upto -∞ towards the left and upto +∞ towards the right.
• The graph is larger in height because:
    ♦ Depending upon the branch, values upto +∞ or –∞ can be obtained as output values.
• The green curve is related to the $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right] - \{ 0 \}$ branch.
• The cyan curve is related to the $\left[\frac{\pi}{2}, \frac{3 \pi}{2} \right] - \{\pi \}$ branch. 
• The magenta curve is related to the $\left[\frac{- 3 \pi}{2}, \frac{- \pi}{2} \right] - \{- \pi \}$ branch.

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In the next section, we will see sec^-1 function.

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18.1 - Inverse of cosine Function

In the previous section, we saw sin^-1 function. In this section, we will see cos^-1 function.

Some basics can be written in 8 steps:
1. Consider the cosine function:
f(x) = cos x
• The input x can be any real number. So the domain is R.
• The output lies in the interval [-1,1]. So the codomain is [-1,1]
• We saw the above details in class 11. We saw a neat pictorial representation of the above details in the graph of the cosine function. It is shown again in fig.18.6 below:

Fig.18.6

2. Let us check whether the cosine function is one-one.
• Let input x = $\frac{\pi}{2}$.
    ♦ Then the output will be $f \left(\frac{\pi}{2} \right)~=~\cos \left(\frac{\pi}{2} \right)~=~0 $
• Let input x = $\frac{3 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{3 \pi}{2} \right)~=~\cos \left(\frac{3 \pi}{2} \right)~=~0 $   
• Let input x = $\frac{-3 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{-3 \pi}{2} \right)~=~\cos \left(\frac{-3 \pi}{2} \right)~=~0 $

(We can cross check with the graph and confirm that the above inputs and outputs are correct)

• We see that, more than one input values from the domain can give the same output. So the cosine function is not a one-one function.

3. Suppose that, we restrict the input values.
• That is, we take input values only from the set $\left[0, \pi \right]$.
• The codomain is the usual [-1,1]
• Then we will get the green curve shown in fig.18.7 below:

Fig.18.7

• In the green portion, no two inputs will give the same output. So the green portion represents a function which is one-one.

4. Next, we have to prove that, the green portion is onto.
• For that, we can consider any y value from the codomain [-1,1].
• There will be always a x value in $\left[0, \pi \right]$, which will satisfy the equation y = cos x.
• So the green portion is onto.
5. We see that, the green portion is both one-one and onto.
• We can represent this function in the mathematical way:
$\text{cosine}:~ \left[0, \pi \right]~\to~[-1,1]$, defined as f(x) = cos x.
6. If a function is both one-one and onto, the codomain is same as range.
• So we can write:
For this function, the domain is $\left[0, \pi \right]$ and range is [-1,1]
7. We know that, if a function is one-one and onto, it will be invertible. We have seen the properties of inverse functions. Let us apply those properties to our present case. It can be written in 4 steps:
(i) If y = f(x) = cos x is invertible, then there exists a function g such that:
g(y) = x
(ii) The function g will also be one-one and onto.
(iii) The domain of f will be the range of g. So the range of g is $\left[0, \pi \right]$.
(iv) The range of f will be the domain of g. So the domain of g is [-1,1]
(iv) The inverse of sine function is denoted as cos^-1. So we can define the inverse function as:
$\cos^{-1}:~ [-1,1]~\to~\left[0, \pi \right]$, defined as x = g(y) = cos^-1 y.
8. Let us see an example:
• Suppose that, for the inverse function, the input y is $\frac{1}{2}$
• Then we get an equation: $x ~=~\cos^{-1} \left(\frac{1}{2} \right)$
• Our aim is to find x. It can be done in 4 steps:
(i) $x ~=~\cos^{-1} \left(\frac{1}{2} \right)$ is an equation of the form $x ~=~f(y)~=~\cos^{-1} \left(y \right)$
(ii) This is an inverse trigonometric function, where input y = $\frac{1}{2}$.
• Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~\cos x$
• In our present case, it is:
$y~=~\frac{1}{2}~=~\cos x$
(iii) So we have a trigonometric equation:
$\cos x~=~\frac{1}{2}$
• When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11.
• In the present case, we do not need to write many steps. We already know that, $\cos \left(\frac{\pi}{3} \right)~=~\frac{1}{2}$
• So we can write:
$x ~=~\frac{\pi}{3}$

---

The above 8 steps help us to understand the basics about cos^-1 function. Now we will see a few more details. It can be written in 5 steps:
1. We saw that, the cos function is not a one-one function. But to make it one-one, we restricted the domain to $\left[0, \pi \right]$.
2. There are other possible “restricted domains” available.
• $\left[- \pi, 0 \right]$ is shown in magenta color in fig.18.8 below:

Fig.18.8

• $\left[\pi, 2 \pi \right]$ is shown in cyan color in fig,18.8 above.
3. There are infinite number of such restricted domains possible.
• We say that:
    ♦ Each restricted domain gives a corresponding branch of the cos^-1 function.
    ♦ The restricted domain $\left[0, \pi \right]$ gives the principal branch of the cos^-1 function.

4. In class 11, we plotted the cosine function. Now we will plot the inverse. It can be done in 4 steps:
(i) Write the set for the original function f. It must contain a convenient number of ordered pairs.
• $\left(\frac{\pi}{3} , \frac{1}{2}  \right)$ is an example of the ordered pairs in f.
(To get a smooth curve, we must write a large number of ordered pairs)
(ii) Based on set f, we can write set g.
This is done by picking each ordered pair from f and interchanging the positions.
• For example, the point $\left(\frac{\pi}{3} , \frac{1}{2}  \right)$ in the set f will become $\left(\frac{1}{2} , \frac{\pi}{3}  \right)$ in set g.
• Thus we will get the required number of ordered pairs in g.
(iii) Mark each ordered pair of g on the graph paper.
• The first coordinate should be marked along the x-axis.
• The second coordinate should be marked along the y-axis.
(iv) Once all the ordered pairs are marked, draw a smooth curve connecting all the marks.
• The smooth curve is the required graph. It is shown in fig.18.9 below:

Fig.18.9

(v) We see that, the graph is smaller in width but larger in height.
• The graph is smaller in width because:
    ♦ Values to the left of -1 cannot be used as input values.    
    ♦ Also, values to the right of 1 cannot be used as input values.
• The graph is larger in height because:
    ♦ Depending upon the branch, values upto +∞ or –∞ can be obtained as output values.
• The red curve is related to the $\left[0, \pi \right]$ branch.
• The cyan curve is related to the $\left[\pi, 2 \pi \right]$ branch. 
• The magenta curve is related to the $\left[- \pi, 0 \right]$ branch.

5. The cos^-1 function is also known as arc cosine function.

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In the next section, we will see cosec^-1 function.

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Chapter 18 - Inverse Trigonometric Functions

In the previous section, we completed a discussion on relations and functions. In this chapter, we will see inverse trigonometric functions.

Some basics can be written in 8 steps:
1. Consider the sine function:
f(x) = sin x
• The input x can be any real number. So the domain is R.
• The output lies in the interval [-1,1]. So the codomain is [-1,1]
• We saw the above details in class 11. We saw a neat pictorial representation of the above details in the graph of the sine function. It is shown again in fig.18.1 below:

Fig.18.1

2. Let us check whether the sine function is one-one.
• Let input x = $\frac{\pi}{2}$.
    ♦ Then the output will be $f \left(\frac{\pi}{2} \right)~=~\sin \left(\frac{\pi}{2} \right)~=~1 $
• Let input x = $\frac{5 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{5 \pi}{2} \right)~=~\sin \left(\frac{5 \pi}{2} \right)~=~1 $   
• Let input x = $\frac{-3 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{-3 \pi}{2} \right)~=~\sin \left(\frac{-3 \pi}{2} \right)~=~1 $

(We can cross check with the graph and confirm that the above inputs and outputs are correct)

• We see that, more than one input values from the domain can give the same output. So the sine function is not a one-one function.

3. Suppose that, we restrict the input values.
• That is, we take input values only from the set $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$.
• The codomain is the usual [-1,1]
• Then we will get the green curve shown in fig.18.2 below:

Fig.18.2

• In the green portion, no two inputs will give the same output. So the green portion represents a function which is one-one.

4. Next, we have to prove that, the green portion is onto.
• For that, we can consider any y value from the codomain [-1,1].
• There will be always a x value in $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$, which will satisfy the equation y = sin x.
• So the green portion is onto.
5. We see that, the green portion is both one-one and onto.
• We can represent this function in the mathematical way:
$\text{sine}:~ \left[\frac{-\pi}{2}, \frac{\pi}{2} \right]~\to~[-1,1]$, defined as f(x) = sin x.
6. If a function is both one-one and onto, the codomain is same as range.
• So we can write:
For this function, the domain is $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$ and range is [-1,1]
7. We know that, if a function is one-one and onto, it will be invertible. We have seen the properties of inverse functions. Let us apply those properties to our present case. It can be written in 4 steps:
(i) If y = f(x) = sin x is invertible, then there exists a function g such that:
g(y) = x
(ii) The function g will also be one-one and onto.
(iii) The domain of f will be the range of g. So the range of g is $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$.
(iv) The range of f will be the domain of g. So the domain of g is [-1,1]
(iv) The inverse of sine function is denoted as sin^-1. So we can define the inverse function as:
$\sin^{-1}:~ [-1,1]~\to~\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$, defined as x = g(y) = sin^-1 y.
8. Let us see an example:
• Suppose that, for the inverse function, the input y is $\frac{1}{2}$
• Then we get an equation: $x ~=~\sin^{-1} \left(\frac{1}{2} \right)$
• Our aim is to find x. It can be done in 4 steps:
(i) $x ~=~\sin^{-1} \left(\frac{1}{2} \right)$ is an equation of the form $x ~=~f(y)~=~\sin^{-1} \left(y \right)$
(ii) This is an inverse trigonometric function, where input y = $\frac{1}{2}$.
• Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~\sin x$
• In our present case, it is:
$y~=~\frac{1}{2}~=~\sin x$
(iii) So we have a trigonometric equation:
$\sin x~=~\frac{1}{2}$
• When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11.
• In the present case, we do not need to write many steps. We already know that, $\sin \left(\frac{\pi}{6} \right)~=~\frac{1}{2}$
• So we can write:
$x ~=~\frac{\pi}{6}$

---

The above 8 steps help us to understand the basics about inverse trigonometric functions. Now we will see a few more details. It can be written in 6 steps:
1. We saw that, the sine function is not a one-one function. But to make it one-one, we restricted the domain to $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]$.
2. There are other possible “restricted domains” available.
• $\left[\frac{-3 \pi}{2}, \frac{- \pi}{2} \right]$ is shown in magenta color in fig.18.3 below:

Fig.18.3

• $\left[\frac{\pi}{2}, \frac{3 \pi}{2} \right]$ is shown in cyan color in fig,18.3 above.
3. There are infinite number of such restricted domains possible.
• We say that:
    ♦ Each restricted domain gives a corresponding branch of the sin^-1 function.
    ♦ The restricted domain $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]$ gives the principal branch of the  sin^-1 function.

4. In class 11, we plotted the sine function. Now we will plot the inverse. It can be done in 4 steps:
(i) Write the set for the original function f. It must contain a convenient number of ordered pairs.
• $\left(\frac{\pi}{6} , \frac{1}{2}  \right)$ is an example of the ordered pairs in f.
(To get a smooth curve, we must write a large number of ordered pairs)
(ii) Based on set f, we can write set g.
This is done by picking each ordered pair from f and interchanging the positions.
• For example, the point $\left(\frac{\pi}{6} , \frac{1}{2}  \right)$ in the set f will become $\left(\frac{1}{2} , \frac{\pi}{6}  \right)$ in set g.
• Thus we will get the required number of ordered pairs in g.
(iii) Mark each ordered pair of g on the graph paper.
• The first coordinate should be marked along the x-axis.
• The second coordinate should be marked along the y-axis.
(iv) Once all the ordered pairs are marked, draw a smooth curve connecting all the marks.
• The smooth curve is the required graph. It is shown in fig.18.4 below:

Fi.18.4

(v) We see that, the graph is smaller in width but larger in height.
• The graph is smaller in width because:
    ♦ Values to the left of -1 cannot be used as input values.    
    ♦ Also, values to the right of 1 cannot be used as input values.
• The graph is larger in height because:
    ♦ Depending upon the branch, values upto +∞ or –∞ can be obtained as output values.
• The green curve is related to the $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$ principal branch.
• The cyan curve is related to the $\left[\frac{\pi}{2}, \frac{3 \pi}{2} \right]$ branch. 
• The magenta curve is related to the $\left[\frac{-3 \pi}{2}, \frac{\pi}{2} \right]$ branch.

5. Consider any function f for which the inverse g exists.
• The graph of g will be the mirror image of the graph of f.
   ♦ The line with equation y=x will be the mirror line.
• We will prove this in the case of sine function. It can be done in 5 steps:
(i) Consider the grid of the graph in fig.18.4 above.
• The x-axis is divided into equal parts, with each part equal to 1 unit.
• The y-axis is also divided into equal parts, with each part equal to π/2 units.
• π/2 = 3.14/2 = 1.57, which is larger than 1. This is the reason why we see larger divisions along the y-axis.
• For the next graph in fig.18.5 below, we will use π/2 for both the axes.

Fig.18.5

(ii) In fig.18.5 above,
    ♦ The sine function is drawn in red color.
    ♦ The sin^-1 function is drawn in green color.
    ♦ The mirror line f(x) = x is drawn in white color.
(iii) Any two convenient points P and Q are marked on the sine function.
• We need to show that, their mirror images P' and Q' lie on the sin^-1 function.
(iv) First we consider point P.
• From P, drop a perpendicular to the mirror line. Let P₁ be the foot of the perpendicular.
• Extend PP₁ so as to intersect with the green curve. Let P' be the point of intersection.
• We can measure and see that, PP₁ = P₁P'.
• Here we see two facts:
    ♦ PP' is perpendicular to the mirror line.
    ♦ PP₁ = P₁P'.
So P' is the mirror image of P.
• We also see that, the coordinates of P' is the interchanged version of P.
(v) Next we consider point Q.
• From Q, drop a perpendicular to the mirror line. Let Q₁ be the foot of the perpendicular.
• Extend QQ₁ so as to intersect with the green curve. Let Q' be the point of intersection.
• We can measure and see that, QQ₁ = Q₁Q'.
• Here we see two facts:
    ♦ QQ' is perpendicular to the mirror line.
    ♦ QQ₁ = Q₁Q'.
So Q' is the mirror image of Q.
• We also see that, the coordinates of Q' is the interchanged version of Q.
• In this way, we can prove that, any point on sin^-1 function is the mirror image of a corresponding point in the sine function.

6. The sin^-1 function is also known as arcsine function.

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We have completed a basic discussion on the sin^-1 function. In the next section, we will see cos^-1 function.

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17.8 - Properties of Invertible Functions

In the previous section, we saw two solved examples related to invertible functions. We saw composite of composite functions also. In this section, we will see two more solved examples. We will also see a new property of invertible functions.

Solved example 17.27
Consider two functions:

f: {1,2,3}→{a,b,c} defined as:
f(1) = a, f(2) = b, f(3) = c

g: {a,b,c}→{apple, ball, cat} defined as:
g(a) = apple, g(b) = ball, g(c) = cat.

I. Show that, f, g and (g∘f) are invertible.
II. Find out f^-1, g^-1 and (g∘f)^-1.
III. Show that (g∘f)^-1 = f^-1∘g^-1

Solution:
Part (I): Proving that f, g and (g∘f) are invertible.
1. f is a function. Domain is {1,2,3} and codomain is {a,b,c}.
• We know that f is a set. The elements of this set are ordered pairs of the form (x,y).
• Based on the given data, we can write all the ordered pairs in f:
f = {(1,a), (2,b), (3,c)}
• We see that:
   ♦ Each element in the domain has a unique image. So f is one-one.
   ♦ All elements in the codomain are images. So f is onto.
• Since f is both one-one and onto, it is invertible.

2. g is a function. Domain is {a,b,c} and codomain is {apple, ball, cat}.
• We know that g is a set. The elements of this set are ordered pairs of the form (x,y).
• Based on the given data, we can write all the ordered pairs in g:
g = {(a,apple), (b,ball), (c,cat)}
• We see that:
   ♦ Each element in the domain has a unique image. So g is one-one.
   ♦ All elements in the codomain are images. So g is onto.
• Since g is both one-one and onto, it is invertible.

3. (g∘f)(x) = g(f(x))
• For (g∘f),
   ♦ domain is the domain of f which is {1,2,3}
   ♦ codomain is the codomain of g which is {apple, ball, cat}
• So we get:
(g∘f)(1) = g(f(1)) = g(a) = apple
(g∘f)(2) = g(f(2)) = g(b) = ball
(g∘f)(3) = g(f(3)) = g(c) = cat

• We know that (g∘f) is a set. The elements of this set are ordered pairs of the form (x,y).
• Based on the above results, we can write all the ordered pairs in (g∘f):
(g∘f) = {(1,apple), (2,ball), (3,cat)}
• We see that:
   ♦ Each element in the domain has a unique image. So (g∘f) is one-one.
   ♦ All elements in the codomain are images. So (g∘f) is onto.
• Since (g∘f) is both one-one and onto, it is invertible.

Part (II): Finding f^-1, g^-1 and (g∘f)^-1.
1. Finding f^-1:
(i) In part (I), we saw that:
f = {(1,a), (2,b), (3,c)}
• So we can write the reverse of f. We will call it p.
• Since f is one-one and onto, the reverse can be easily written:
p = {(a,1), (b,2), (c,3)}
• The domain of p is {a,b,c}. And codomain is {1,2,3}

(ii) Now we have both f and p. We can calculate (p∘f)
(p∘f) means, output of f is used as the input of p.
• The two functions are:
f: {1,2,3}→{a,b,c}
p: {a,b,c}→{1,2,3}
• We know that, (p∘f) directly connects the domain of f to the codomain of p. So (p∘f) is from {1,2,3} to {1,2,3}.
• When the input is 1, we get:
(p∘f)(1) = p(f(1)) = p(a) = 1
• When the input is 2, we get:
(p∘f)(2) = p(f(2)) = p(b) = 2
• When the input is 3, we get:
(p∘f)(3) = p(f(3)) = p(c) = 3

• So whatever input we give, the output will be the same. That means, (p∘f) is an identity function.
• The inputs are taken from {1,2,3}. So we can write:
(p∘f) = I_{1,2,3}.

(iii) Similarly, we can calculate (f∘p)
(f∘p) means, output of p is used as the input of f.
• The two functions are:
f: {1,2,3}→{a,b,c}
p: {a,b,c}→{1,2,3}
• We know that, (f∘p) directly connects the domain of p to the codomain of f. So (f∘p) is from {a,b,c} to {a,b,c}.
• When the input is a, we get:
(f∘p)(a) = f(p(a)) = f(1) = a
• When the input is 2, we get:
(f∘p)(b) = f(p(b)) = f(2) = b
• When the input is 3, we get:
(f∘p)(c) = f(p(c)) = f(3) = c

• So whatever input we give, the output will be the same. That means, (f∘p) is an identity function.
• The inputs are taken from {a,b,c}. So we can write:
(f∘p) = I_{a,b,c}.

(iv). Let us write a summary:
• We are given a function f: {1,2,3}→{a,b,c}
• We wrote the reverse function p: {a,b,c}→{1,2,3}
• From (ii), we got: (p∘f) = I_{1,2,3}
• From (iii), we got: (f∘p) = I_{a,b,c}
• So f is invertible.
• Also, the inverse of f = f^-1 = p
• That means:
The inverse of f: {1,2,3}→{a,b,c} is:
p: {a,b,c}→{1,2,3}

2. Finding g^-1:
(i) In part (I), we saw that:
g = {(a,apple), (b,ball), (c,cat)}
• So we can write the reverse of g. We will call it q.
• Since g is one-one and onto, the reverse can be easily written:
q = {(apple,a), (ball,b), (cat,c)}
• The domain of q is {apple, ball, cat}. And codomain is {a,b,c}

(ii) Now we have both g and q. We can calculate (q∘g)
(q∘g) means, output of g is used as the input of q.
• The two functions are:
g: {a,b,c}→{apple, ball, cat}
q: {apple, ball, cat}→{a,b,c}
• We know that, (q∘g) directly connects the domain of g to the codomain of q. So (q∘g) is from {a,b,c} to {a,b,c}.
• When the input is a, we get:
(q∘g)(a) = q(g(a)) = q(apple) = a
• When the input is b, we get:
(q∘g)(b) = q(g(b)) = q(ball) = b
• When the input is c, we get:
(q∘g)(c) = q(g(c)) = q(cat) = c

• So whatever input we give, the output will be the same. That means, (q∘g) is an identity function.
• The inputs are taken from {a,b,c}. So we can write:
(q∘g) = I_{a,b,c}.

(iii) Similarly, we can calculate (g∘q)
(g∘q) means, output of q is used as the input of g.
• The two functions are:
g: {a,b,c}→{apple, ball, cat}
q: {apple, ball, cat}→{a,b,c}
• We know that, (g∘q) directly connects the domain of q to the codomain of g. So (g∘q) is from {apple, ball, cat} to {apple, ball, cat}.
• When the input is apple, we get:
(g∘q)(apple) = g(q(apple)) = g(a) = apple
• When the input is 2, we get:
(g∘q)(ball) = g(q(ball)) = g(b) = ball
• When the input is 3, we get:
(g∘q)(cat) = g(q(cat)) = g(c) = cat

• So whatever input we give, the output will be the same. That means, (g∘q) is an identity function.
• The inputs are taken from {apple, ball, cat}. So we can write:
(g∘q) = I_{{apple, ball, cat}}.

(iv). Let us write a summary:
• We are given a function g: {a,b,c}→{apple, ball, cat}
• We wrote the reverse function q: {apple, ball, cat}→{a,b,c}
• From (ii), we got: (q∘g) = I_{a,b,c}
• From (iii), we got: (g∘q) = I_{{apple, ball, cat}}
• So g is invertible.
• Also, the inverse of g = g^-1 = q
• That means:
The inverse of g: {a,b,c}→{apple, ball, cat} is:
q: {apple, ball, cat}→{a,b,c}

3. Finding (g∘f)^-1:
(i) In part (I), we saw that:
(g∘f) = {(1,apple), (2,ball), (3,cat)}
• So we can write the reverse of (g∘f). We will call it r.
• Since (g∘f) is one-one and onto, the reverse can be easily written:
r = {(apple,1), (ball,2), (cat,3)}
• The domain of r is {apple, ball, cat}. And codomain is {1,2,3}

(ii) Now we have both (g∘f) and r. We can calculate (r∘(g∘f))
(r∘(g∘f)) means, output of (g∘f) is used as the input of r.
• The two functions are:
(g∘f): {1,2,3}→{apple, ball, cat}
r: {apple, ball, cat}→{1,2,3}
• We know that, (r∘(g∘f)) directly connects the domain of (g∘f) to the codomain of r. So (r∘(g∘f)) is from {1,2,3} to {1,2,3}.
• When the input is 1, we get:
(r∘(g∘f))(1) = r((g∘f)(1)) = r(apple) = 1
• When the input is 2, we get:
(r∘(g∘f))(2) = r((g∘f)(2)) = r(ball) = 2
• When the input is 3, we get:
(r∘(g∘f))(3) = r((g∘f)(3)) = r(cat) = 3

• So whatever input we give, the output will be the same. That means, (r∘(g∘f)) is an identity function.
• The inputs are taken from {1,2,3}. So we can write:
(r∘(g∘f)) = I_{1,2,3}.

(iii) Similarly, we can calculate ((g∘f)∘r)
((g∘f)∘r) means, output of r is used as the input of (g∘f).
• The two functions are:
(g∘f): {1,2,3}→{apple, ball, cat}
r: {apple, ball, cat}→{1,2,3}
• We know that, ((g∘f)∘r) directly connects the domain of r to the codomain of (g∘f). So ((g∘f)∘r) is from {apple, ball, cat} to {apple, ball, cat}.
• When the input is apple, we get:
((g∘f)∘r)(apple) = (g∘f)(r(apple)) = (g∘f)(1) = apple
• When the input is 2, we get:
((g∘f)∘r)(ball) = (g∘f)(r(ball)) = (g∘f)(2) = ball
• When the input is 3, we get:
((g∘f)∘r)(cat) = (g∘f)(r(cat)) = (g∘f)(3) = cat

• So whatever input we give, the output will be the same. That means, ((g∘f)∘r) is an identity function.
• The inputs are taken from {apple, ball, cat}. So we can write:
((g∘f)∘r) = I_{{apple, ball, cat}}.

(iv). Let us write a summary:
• We are given a function (g∘f): {1,2,3}→{apple, ball, cat}
• We wrote the reverse function r: {apple, ball, cat}→{1,2,3}
• From (ii), we got: (r∘(g∘f)) = I_{1,2,3}
• From (iii), we got: ((g∘f)∘r) = I_{{apple, ball, cat}}
• So (g∘f) is invertible.
• Also, the inverse of (g∘f) = (g∘f)^-1 = r
• That means:
The inverse of (g∘f): {1,2,3}→{apple, ball, cat} is:
r: {apple, ball, cat}→{1,2,3}

Part III: Showing that, (g∘f)^-1 = f^-1∘g^-1.
1. From part (II)(1), we have:
f^-1 = p
Where p: {a,b,c}→{1,2,3}
• In set form, we can write:
f^-1 = {(a,1), (b,2), (c,3)}


2. From part (II)(2), we have:
g^-1 = q
Where q: {apple, ball, cat}→{a,b,c}
• In set form, we can write:
q^-1 = {(apple,a), (ball,b), (cat,c)}

3. From part (II)(3), we have:
(g∘f)^-1 = r
Where r: {apple, ball, cat}→{1,2,3}
• In set form, we can write:
(g∘f)^-1 = {(apple,1), (ball,2), (cat,3)}

4. Now we can calculate f^-1∘g^-1.
(f^-1∘g^-1)(x) = f^-1(g^-1(x))
• For (f^-1∘g^-1),
   ♦ domain is the domain of g^-1 which is {apple, ball, cat}
   ♦ codomain is the codomain of f^-1 which is {1,2,3}
• So we get:
(f^-1∘g^-1)(apple) = f^-1(g^-1(apple)) = f^-1(a) = 1
(f^-1∘g^-1)(ball) = f^-1(g^-1(ball)) = f^-1(b) = 2
(f^-1∘g^-1)(cat) = f^-1(g^-1(cat)) = f^-1(c) = 3

5. We can write the result in (4), in the form of a set:
(f^-1∘g^-1) = {(apple,1), (ball,2), (cat,3)}

6. Comparing the results in (3) and (5), we get:
(g∘f)^-1 = (f^-1∘g^-1)

---

The result obtained in the above solved example 17.27 can be used in general. We can write:
If f: X→Y and g: Y→Z are two invertible functions, then (g∘f) is also invertible. The inverse of (g∘f) is (f^-1∘g^-1)

The proof can be written in 7 steps:
1. First we will write some basic details:
(i) Given that, f: X→Y is an invertible function.
• So we can write:
    ♦ f(x) = y
    ♦ f^-1(y) = x
(ii) Given that, g: Y→Z is an invertible function.
• So we can write:
    ♦ g(y) = z
    ♦ g^-1(z) = y
(iii) Note that, domain of g is same as codomain of f.

2. Now we recall an important property of inverse functions.
• If the inverse of f is f^-1, then two conditions will be satisfied:
(i) (f^-1∘f) = I_X
(ii) (f∘f^-1) = I_Y
Where
    ♦ X is the domain of f
    ♦ Y is the domain of f^-1 

• Similarly, if the inverse of g is g^-1, then two conditions will be satisfied:
(i) (g^-1∘g) = I_Y
(ii) (g∘g^-1) = I_Z
Where
    ♦ Y is the domain of g
    ♦ Z is the domain of g^-1

3. Next step is to determine the domain and codomain of the composite functions.
(i) Domain of (g∘f):
• (g∘f) connects the domain of f with the codomain of g.
    ♦ So the domain of (g∘f) is the domain of f, which is X.
    ♦ Also the codomain of (g∘f) is the codomain of g, which is Y. 
(ii) Domain of (f^-1∘g^-1):
• (f^-1∘g^-1) connects the domain of g^-1 with the codomain of f^-1.
    ♦ So the domain of (f^-1∘g^-1) is the domain of g^-1, which is Z.
    ♦ Also the codomain of (f^-1∘g^-1) is the codomain of f^-1, which is X. 

4. Now we can use the property mentioned in step (2).
• That is:
If the inverse of (g∘f) is (f^-1∘g^-1), then two conditions will be satisfied:
(i) (f^-1∘g^-1)∘(g∘f) = I_X
(ii) (g∘f)∘(f^-1∘g^-1) = I_Z
Where
    ♦ X is the domain of (g∘f)
    ♦ Z is the domain of (f^-1∘g^-1)

5. First we show that (f^-1∘g^-1)∘(g∘f) = I_X.
$\begin{array}{ll}{}    &{(f^{-1} \circ g^{-1})(g \circ f)}    & {~=~}    &{\Bigl((f^{-1} \circ g^{-1}) \circ g \Bigr) \circ f ~\color {magenta}{\text{- - - (A)}}}    &{} \\
{}    &{}    & {~=~}    &{\Bigl(f^{-1} \circ (g^{-1} \circ g) \Bigr) \circ f ~\color {magenta}{\text{- - - (B)}}}    &{} \\
{}    &{}    & {~=~}    &{(f^{-1} \circ I_{Y}) \circ f ~\color {magenta}{\text{- - - (C)}}}    &{} \\
{}    &{}    & {~=~}    &{I_X ~\color {magenta}{\text{- - - (D)}}}    &{} \\
\end{array}               
$

◼ Remarks:
• Line marked as A:
Here we use the formula: h∘(g∘f) = (h∘g)∘f
• Line marked as B:
Here also, we use the formula: h∘(g∘f) = (h∘g)∘f
• Line marked as C:
Here we use the fact that: (g^-1∘g) = I_Y
• Line marked as D:
We must simplify (f^-1∘I_Y)∘f. This can be done in two steps:
(i) First we will determine (f^-1∘I_Y)
• Here, the output of I_Y is used as the input for f^-1.
• Also this function connects the domain of I_Y with the codomain of f^-1.
    ♦ Domain of I_Y is Y
    ♦ Codomain of f^-1 is X
So we get: (f^-1∘I_Y) = f^-1(I_Y) = f^-1(y) = x.
(ii) Now we can determine (f^-1∘I_Y)∘f
• Here, the output of f is used as the input for (f^-1∘I_Y).
• Also this function connects the domain of f with the codomain of (f^-1∘I_Y).
    ♦ Domain of f is X
    ♦ Codomain of (f^-1∘I_Y) is also X
So we get: (f^-1∘I_Y)∘f = I_X.

6. In a similar way, we can show that:
(g∘f)∘(f^-1∘g^-1) = I_Z.

7. So both conditions mentioned in (4) are satisfied.
• Therefore it is proved that:
If f: X→Y and g: Y→Z are two invertible functions, then (g∘f) is also invertible. The inverse of (g∘f) is (f^-1∘g^-1)  

Solved example 17.28
Let S = {1,2,3}. Determine whether the functions f: S→S defined as below have inverses. Find f^-1 if it exists.
(a) f = {(1,1), (2,2), (3,3)}
(b) f = {(1,2), (2,1), (3,1)}
(c) f = {(1,3), (3,2), (2,1)}
Solution:
Part (a): f = {(1,1), (2,2), (3,3)}
• Here f is both one-one and onto. So f^-1 exists.
• We can write: f^-1 = {(1,1), (2,2), (3,3)}

Part (b): f = {(1,2), (2,1), (3,1)}
• Here f(2) = f(3) = 1.
So f is not one-one and onto. Therefore, f^-1 does not exist.

Part (c): f = {(1,3), (3,2), (2,1)}
• Here f is both one-one and onto. So f^-1 exists.
• We can write: f^-1 = {(3,1), (2,3), (1,2)}

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Link to a few more solved examples is given below:

Exercise 17.3

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In the next section, we will see binary operations.

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