Inverse Trigonometric Functions

Inverse Trigonometric Functions in Mathematics

Inverse Trigonometric Functions in Mathematics help in finding angles when the values of trigonometric ratios are known. They play a crucial role in calculus and have numerous applications in physics, engineering, and higher mathematics.

What are Inverse Trigonometric Functions?

Inverse trigonometric functions enable us to find angles when the value of a trigonometric ratio is given. They play an important role in solving problems related to geometry, calculus, and real-life applications.

Inverse trigonometric functions are used to determine an angle from a given trigonometric value. For example, if $\sin \theta = \frac{1}{2}$, then $\theta = \sin^{-1}(\frac{1}{2})$. In inverse trigonometric functions class 12, you will learn how to apply inverse trigonometric functions formulas, understand the domain and range of inverse trigonometric functions, and interpret the graph of inverse trigonometric functions.

Inverse trigonometric functions are the reverse process of the trigonometric functions. In other words, the domain of the inverse function is the range of the original function and vice versa.

For example:

1. $\sin(\frac{\pi}{6}) = \frac{1}{2}$ ⇒ $\frac{\pi}{6} = \sin^{-1}(\frac{1}{2})$

2. $\cos(\pi) = -1$ ⇒ $\pi = \cos^{-1}(-1)$

3. $\tan(\frac{\pi}{4}) = 1$ ⇒ $\frac{\pi}{4} = \tan^{-1}(1)$

Types of Inverse Trigonometric Functions

1. Arcsine ($\sin^{-1}$) – Inverse of the sine function

2. Arccosine ($\cos^{-1}$) – Inverse of the cosine function

3. Arctangent ($\tan^{-1}$) – Inverse of the tangent function

4. Arccotangent ($\cot^{-1}$) – Inverse of the cotangent function

5. Arcsecant ($\sec^{-1}$) – Inverse of the secant function

6. Arccosecant ($\csc^{-1}$) – Inverse of the cosecant function

These functions are covered in the inverse trigonometric functions class 12 and include important formulas, domain and range explanations, and graphs.

Domain and Range of Inverse Trigonometric Functions

Understanding the domain and range of inverse trigonometric functions is essential for solving problems. Below are the domains and ranges for each inverse function.

To define inverse trigonometric functions properly, the original trigonometric functions must be one-one and onto in the selected domain and range. For example, the sine function is defined for all real numbers, but its inverse is only defined when the domain is restricted to $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and the range to $[-1, 1]$.

Thus,

• The domain of $y = \sin^{-1}(x)$ is $[-1, 1]$

• The range of $y = \sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$

The same process applies to other functions like cosine, tangent, secant, etc.

Principal Domain / Principal Value Branch

By convention, the domain of inverse trigonometric functions is chosen to ensure the function is one-one and onto. For example, the principal domain for $y = \sin(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, which makes $y = \sin^{-1}(x)$ well-defined.

Graph of Inverse Trigonometric Functions

The graphs of inverse trigonometric functions are given below:

$y=\sin ^{-1}(x)$

$\sin ^{-1}(x)$ is the inverse of the trignometric function $\sin x$.

$\mathrm{y}=\sin \mathrm{x}, \mathrm{x} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\mathrm{y} \in[-1,1]$

$y=\cos ^{-1}(x)$

$\cos ^{-1}(x)$ is the inverse of $\cos x$

$y=\tan ^{-1}(x)$

$\tan ^{-1}(x)$ is the inverse of $\tan x$.

$y=\operatorname{cosec}^{-1}(x)$

$\operatorname{cosec}^{-1}(x)$ is the inverse of $\operatorname{cosec} x$.

$y=\sec ^{-1}(x)$

$\sec ^{-1}(x)$ is the inverse of $\sec x$.

$y=\cot ^{-1}(x)$

$\cot ^{-1}(x)$ is the inverse of $\cot x$.

Inverse Trigonometric Functions Formulas

This section covers important inverse trigonometric functions formulas, including formulas for negative functions, reciprocal functions, complementary functions, sum and difference of angles, multiple angles, as well as differentiation and integration formulas. These formulas are essential for inverse trigonometric functions class 12 and are helpful in solving calculus problems.

Formulas for Negative Functions

• $\sin^{-1}(-x) = -\sin^{-1}(x)$, for all $x \in [-1,1]$

• $\tan^{-1}(-x) = -\tan^{-1}(x)$, for all $x \in \mathbb{R}$

• $\csc^{-1}(-x) = -\csc^{-1}(x)$, for all $x \in \mathbb{R} - (-1,1)$

• $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$, for all $x \in [-1,1]$

• $\sec^{-1}(-x) = \pi - \sec^{-1}(x)$, for all $x \in \mathbb{R} - (-1,1)$

• $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$, for all $x \in \mathbb{R}$

Formulas for Reciprocal Functions

• $\sin^{-1}\left(\frac{1}{x}\right) = \csc^{-1}(x)$, for all $x \in (-\infty,-1] \cup [1,\infty)$

• $\cos^{-1}\left(\frac{1}{x}\right) = \sec^{-1}(x)$, for all $x \in (-\infty,-1] \cup [1,\infty)$

• $\tan^{-1}\left(\frac{1}{x}\right) = \begin{cases}\cot^{-1} x, & x>0\ -\pi + \cot^{-1} x, & x<0\end{cases}$

Formulas for Complementary Functions

The sum of complementary functions results in a right angle:

• $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, for all $x \in [-1,1]$

• $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$, for all $x \in \mathbb{R}$

• $\sec^{-1} x + \csc^{-1} x = \frac{\pi}{2}$, for all $x \in (-\infty,-1] \cup [1,\infty)$

Sum and Difference Formulas of Inverse Trigonometric Functions

Below are the sum and difference formulas of inverse trigonometric functions like $\tan^{-1}$, $\sin^{-1}$, and $\cos^{-1}$. These formulas help in simplifying expressions and solving equations in the inverse trigonometric functions class 12.

Sum of Angles in Terms of $\tan^{-1}$

• $\tan^{-1} x + \tan^{-1} y = \begin{cases} \tan^{-1}\left(\frac{x+y}{1-xy}\right), & x>0, y>0, xy<1\\ \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right), & x>0, y>0, xy>1\\ -\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right), & x<0, y<0, xy>1 \end{cases}$

Difference of Angles in Terms of $\tan^{-1}$

• $\tan^{-1} x - \tan^{-1} y = \begin{cases} \tan^{-1}\left(\frac{x-y}{1+xy}\right), & xy > -1\\ \pi + \tan^{-1}\left(\frac{x-y}{1+xy}\right), & x>0, y<0, xy<-1\\ -\pi + \tan^{-1}\left(\frac{x-y}{1+xy}\right), & x<0, y>0, xy<-1 \end{cases}$

Sum and Difference of Angles in Terms of $\sin^{-1}$

• $\sin^{-1} x + \sin^{-1} y = \begin{cases} \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right), & -1 \leq x, y \leq 1, x^2+y^2 \leq 1\\ \pi - \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right), & x>0, y>0, x^2+y^2>1\\ -\pi - \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right), & -1 \leq x, y <0, x^2+y^2>1\end{cases}$

• $\sin^{-1} x - \sin^{-1} y = \begin{cases} \sin^{-1}\left(x\sqrt{1-y^2} - y\sqrt{1-x^2}\right), & -1 \leq x, y \leq 1, x^2+y^2 \leq 1\\ \pi - \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right), & x>0, y<0, x^2+y^2>1\\ -\pi - \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right), & -1 \leq x<0, 0<y\leq1, x^2+y^2>1\end{cases}$

Sum and Difference of Angles in Terms of $\cos^{-1}$

• $\cos^{-1} x + \cos^{-1} y = \cos^{-1}\left(xy - \sqrt{1-x^2}\sqrt{1-y^2}\right)$, if $0 < x, y \leq 1$

• $\cos^{-1} x - \cos^{-1} y = \begin{cases} \cos^{-1}\left(xy + \sqrt{1-x^2}\sqrt{1-y^2}\right), & 0 \leq x, y \leq 1, x \leq y\\ -\cos^{-1}\left(xy + \sqrt{1-x^2}\sqrt{1-y^2}\right), & 0 < x, y \leq 1, x > y \end{cases}$

Multiple Angle Formulas of Inverse Trigonometric Functions

Below are the multiple angle formulas of inverse trigonometric functions such as $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$. These formulas are useful for solving advanced problems in inverse trigonometric functions, Class 12, and calculus.

Double and Triple Angle Formulas in Terms of $\sin^{-1}$

• $2\sin^{-1} x = \begin{cases} \sin^{-1}(2x\sqrt{1-x^2}), & -\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}\\ \pi - \sin^{-1}(2x\sqrt{1-x^2}), & x > \frac{1}{\sqrt{2}}\\ -\pi - \sin^{-1}(2x\sqrt{1-x^2}), & x < -\frac{1}{\sqrt{2}} \end{cases}$

• $3\sin^{-1} x = \begin{cases} \sin^{-1}(3x - 4x^3), & -\frac{1}{2} \leq x \leq \frac{1}{2}\\ \pi - \sin^{-1}(3x - 4x^3), & x > \frac{1}{2}\\ -\pi - \sin^{-1}(3x - 4x^3), & x < -\frac{1}{2}\end{cases}$

Double and Triple Angle Formulas in Terms of $\cos^{-1}$

• $2\cos^{-1} x = \begin{cases} \cos^{-1}(2x^2 - 1), & 0 \leq x \leq 1\\ 2\pi - \cos^{-1}(2x^2 - 1), & -1 \leq x \leq 0\end{cases}$

• $3\cos^{-1} x = \begin{cases} \cos^{-1}(4x^3 - 3x), & \frac{1}{2} \leq x \leq 1\\ 2\pi - \cos^{-1}(4x^3 - 3x), & -\frac{1}{2} \leq x \leq \frac{1}{2}\\ 2\pi + \cos^{-1}(4x^3 - 3x), & -1 \leq x \leq -\frac{1}{2}\end{cases}$

Multiple Angles in Terms of $\tan^{-1}$ and $\sin^{-1}$

• $2\tan^{-1} x = \begin{cases} \sin^{-1}\left(\frac{2x}{1+x^2}\right), & -1 \leq x \leq 1\\ \pi - \sin^{-1}\left(\frac{2x}{1+x^2}\right), & x>1\\ -\pi - \sin^{-1}\left(\frac{2x}{1+x^2}\right), & x<-1\end{cases}$

Multiple Angles in Terms of $\tan^{-1}$ and $\cos^{-1}$

• $2\tan^{-1} x = \begin{cases} \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), & 0 \leq x < \infty\\ -\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), & -\infty < x \leq 0\end{cases}$

Differentiation of Inverse Trigonometric Functions

The derivatives of inverse trigonometric functions are essential for calculus problems:

• $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}, x \neq \pm 1$

• $\frac{d}{dx}(\cos^{-1} x) = \frac{-1}{\sqrt{1-x^2}}, x \neq \pm 1$

• $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$

• $\frac{d}{dx}(\cot^{-1} x) = \frac{-1}{1+x^2}$

• $\frac{d}{dx}(\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2-1}}, x \neq \pm 1$

• $\frac{d}{dx}(\csc^{-1} x) = \frac{-1}{|x|\sqrt{x^2-1}}, x \neq \pm 1$

Integration of Inverse Trigonometric Functions

The integrals of inverse trigonometric functions are widely used in calculus:

• $\int \sin^{-1} x ,dx = x\sin^{-1}x + \sqrt{1-x^2} + C$

• $\int \cos^{-1} x ,dx = x\cos^{-1}x - \sqrt{1-x^2} + C$

• $\int \tan^{-1} x ,dx = x\tan^{-1}x - \frac{1}{2}\ln|1+x^2| + C$

• $\int \csc^{-1} x ,dx = x\csc^{-1}x + \ln|x+\sqrt{x^2-1}| + C$

• $\int \sec^{-1} x ,dx = x\sec^{-1}x - \ln|x+\sqrt{x^2-1}| + C$

• $\int \cot^{-1} x ,dx = x\cot^{-1}x + \frac{1}{2}\ln|1+x^2| + C$

Inverse Trigonometric Functions Table

This table summarises the domain and range of inverse trigonometric functions, which is an important topic for the inverse trigonometric functions class 12 and solving problems.

Difference between Trigonometric Functions and Inverse Trigonometric Functions

Below is a table showing the key differences between trigonometric functions and inverse trigonometric functions, covering their definitions, domains, ranges, and applications in the inverse trigonometric functions class 12.

Inverse Trigonometric Functions in Mathematics: Solved Previous Year Questions

Question 1:

The sum of possible values of $x$ for $\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1}\left(\frac{8}{31}\right)$ is:

Solution:

$
\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1} \frac{8}{31}
$
Taking tangent on both sides:

$
\begin{aligned}
& \frac{(x+1)+(x-1)}{1-\left(x^2-1\right)}=\frac{8}{31} \\
& \Rightarrow \frac{2 x}{2-x^2}=\frac{8}{31} \\
& \Rightarrow 4 x^2+31 x-8=0 \\
& \Rightarrow x=-8, \frac{1}{4}
\end{aligned}
$
But, if $x=\frac{1}{4}$

$
\tan ^{-1}(x+1) \in\left(0, \frac{\pi}{2}\right)
$

$\& \cot ^{-1}\left(\frac{1}{x-1}\right) \in\left(\frac{\pi}{2}, \pi\right)$
$\Rightarrow L H S>\frac{\pi}{2} \quad \& \quad R H S<\frac{\pi}{2}$
(Not possible)
Hence, $x=-8$

Hence, the correct answer is the $-\frac{32}{4}$.

Question 2:

Given that the inverse trigonometric function takes principal values only. Then , the number of real values of x which satisfy $\sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1}\left(\frac{4 x}{5}\right)=\sin ^{-1} x$ is equal to:

Solution:

$
\begin{aligned}
& \sin ^{-1} \frac{3 x}{5}+\sin ^{-1} \frac{4 x}{5}=\sin ^{-1} x \\
& \sin ^{-1}\left(\frac{3 x}{5} \sqrt{1-\frac{16 x^2}{25}}+\frac{4 x}{5} \sqrt{1-\frac{9 x^2}{25}}\right)=\sin ^{-1} x \\
& \frac{3 x}{5} \sqrt{1-\frac{16 x^2}{25}}+\frac{4 x}{5} \sqrt{1-\frac{9 x^2}{25}}=x \\
& x=0,3 \sqrt{25-16 x^2}+4 \sqrt{25-9 x^2}=25 \\
& 4 \sqrt{25-9 x^2}=25-3 \sqrt{25-16 x^2} \text { squaring we get } \\
& 16\left(25-9 x^2\right)=625+9\left(25-16 x^2\right)-150 \sqrt{25-16 x^2} \\
& 400=625+225-150 \sqrt{25-16 x^2} \\
& 400=625+225-150 \sqrt{25-16 x^2} \\
& \sqrt{25-16 x^2}=3 \\
& \Rightarrow 25-16 x^2=9 \\
& \Rightarrow x^2=1
\end{aligned}
$
Put $x=0,1,-1$ in the original equation, we see that all values satisfy the original equation.
Number of solutions $=3$

Hence, the correct answer is 3.

Question 3:

If $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=4 \tan ^{-1} x$, then the number of solutions of this equation is (given $x>1$ )

Solution:

LHS $: \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$
Let $\tan ^{-1} x=\theta \Rightarrow \tan \theta=x$

$
=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)=\tan ^{-1}(\tan 2 \theta)
$

(Now

$
\begin{aligned}
& 1<x<\infty \Rightarrow \frac{\pi}{4}<\tan ^{-1} x<\frac{\pi}{2} \\
& \Rightarrow \frac{\pi}{4}<\theta<\frac{\pi}{2} \Rightarrow \frac{\pi}{2}<2 \theta<\pi
\end{aligned}
$

For this branch, $\tan ^{-1}(\tan x)=x-\pi$

$
\Rightarrow \tan ^{-1}(\tan 2 \theta)=2 \theta-\pi=2 \tan ^{-1} x-\pi
$

Now the equation is

$
\begin{aligned}
& 2 \tan ^{-1} x-\pi=4 \tan ^{-1} x \\
& \Rightarrow 2 \tan ^{-1} x=-\pi \\
& \Rightarrow \tan ^{-1} x=-\frac{\pi}{2}
\end{aligned}
$

which is not possible. So the number of solutions $=0$

Hence, the correct answer is 0.

Question 4:

The solution of $\sin^{-1}x+\sin^{-1}\left ( 2x \right )=\frac{\pi}{2}$ is:

Solution:

$
\begin{aligned}
& \sin ^{-1} x+\sin ^{-1}(2 x)=\frac{\pi}{2} \\
& \Rightarrow \sin ^{-1}(2 x)=\frac{\pi}{2}-\sin ^{-1} x
\end{aligned}
$

Taking sine on both sides
$
\begin{aligned}
& \Rightarrow \sin \left(\sin ^{-1}(2 x)\right)=\sin \left(\frac{\pi}{2}-\sin ^{-1}(x)\right) \\
& \Rightarrow 2 x=\sin \left(\cos ^{-1} x\right) \quad\left(\text { Using } \frac{\pi}{2}-\sin ^{-1} x=\cos ^{-1} x\right) \\
& \Rightarrow 2 x=\sin \left(\sin ^{-1} \sqrt{1-x^2}\right) \quad \text { (using interconversion) } \\
& \Rightarrow 2 x=\sqrt{1-x^2} \\
& \Rightarrow 4 x^2=1-x^2 \\
& \Rightarrow 3 x^2=1
\end{aligned}
$

$
\Rightarrow x=\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}
$

Check:
$
x=\frac{1}{\sqrt{3}}
$

Both $\sin ^{-1}(x)$ and $\sin ^{-1}(2 x)>0$, so they can add up to $\frac{\pi}{2}$. So $\quad x=\frac{1}{\sqrt{3}}$ is correct.
$
x=\frac{-1}{\sqrt{3}}
$

Both terms on $L H S<0 \Rightarrow L H S<0$.
So $\quad x=\frac{-1}{\sqrt{3}}$ is rejected.

Hence, the correct answer is $\frac{1}{\sqrt{3}}$.

Question 5:

The value of $\sum_{r=0}^5 \tan ^{-1}\left(\frac{1}{1+r+r^2}\right)$ is:

Solution:

$\begin{aligned} & \tan ^{-1}\left(\frac{1}{1+r+r^2}\right)=\tan ^{-1}\left(\frac{1}{1+r(r+1)}\right)=\tan ^{-1}\left(\frac{(r+1)-r}{1+r(r+1)}\right) \\ & \quad=\tan ^{-1}(r+1)-\tan ^{-1}(r) \\ & \therefore \sum_{r=0}^5 \tan ^{-1}\left(\frac{1}{1+r+r^2}\right)=\left(\tan ^{-1} 1-\tan ^{-1} 0\right)+\left(\tan ^{-1} 2-\tan ^{-1} 1\right)+ \\ & \quad \ldots \ldots \ldots . .+\left(\tan ^{-1} 5-\tan ^{-1} 4\right)+\left(\tan ^{-1} 6-\tan ^{-1} 5\right) \\ & =\tan ^{-1} 6-\tan ^{-1} 0 \\ & =\tan ^{-1} 6\end{aligned}$

Hence, the correct answer is $\tan ^{-1} 6$.

List of Topics according to NCERT/JEE Mains

Below is a list of important topics related to the NCERT syllabus and JEE Main, helping you focus on key areas for preparation.

• Relations and Functions
• Matrices
• Determinants
• Continuity and Differentiability
• Application of Derivatives
• Integrals

Inverse Trigonometric Functions in Different Exams

The chapter on Inverse Trigonometric Functions is an important part of Class 12 Mathematics and is frequently asked in board exams as well as competitive examinations. It focuses on understanding inverse relations of trigonometric functions and their properties. The table below highlights the exam-wise focus areas, commonly asked topics, and preparation strategies to help students prepare effectively.

Important Books for Inverse Trigonometric Functions

Below are the recommended books for studying inverse trigonometric functions, including theory, examples, and practice questions, useful for inverse trigonometric functions class 12 and competitive exams like JEE Main.

NCERT Resources for Inverse Trigonometric Functions

Below are helpful NCERT resources such as textbooks, solutions, and notes that explain inverse trigonometric functions clearly, making it easier to prepare for board exams and entrance tests.

• NCERT Maths Notes for Class 12th Chapter 2 Inverse Trigonometric Functions
• NCERT Maths Solutions for Class 12th Chapter 2 Inverse Trigonometric Functions
• NCERT Maths Exemplar Solutions for Class 12th Chapter 2 Inverse Trigonometric Functions

NCERT Subjectwise Resources

Below are subject-wise NCERT resources offering notes, solutions, exemplar problems, and solved examples for class 12 and competitive exam preparation.

Practice Questions based on Inverse Trigonometric Functions

Below are practice questions based on inverse trigonometric functions, including problems from NCERT exercises and JEE Main-level questions, to help you strengthen your understanding and problem-solving skills.

Conclusion

Inverse Trigonometric Functions play an important role in finding unknown angles from known trigonometric values. A clear understanding of their definitions, domain and range, formulas, and graphs makes solving problems easier and more accurate. With regular practice and proper revision, students can confidently handle inverse trigonometric function questions in board and competitive examinations.