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:
$x ~=~\frac{\pi}{3}$

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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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