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:
$x ~=~\frac{\pi}{6}$
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.
In the next section, we will see cot-1 function.
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