Inverse Trigonometric Functions: Introduction, Properties and Examples

You have landed on the right page to learn about Inverse Trigonometric Functions. We know that the inverse of a function \(f\), denoted by \(f^{-1}\), exists if \(f\) is one-one and onto. Several functions are not one-one, onto or both, so we cannot know about their inverse. Also, we know that the trigonometric functions are not one-one and onto over their natural domains and ranges, and hence their inverses do not exist in their natural domain.

This article will study the restrictions on domains and ranges of trigonometric functions, ensuring their inverses and observing their behaviour through graphical representation. We will also learn some properties of inverse trigonometric functions.

Inverse Function

Let \(f: X \rightarrow Y\) be a bijective function.

If we can make another function \(g\) from \(Y\) to \(X\), we can say that \(g\) is the inverse of \(f\).

i.e., \(g=f^{-1} \neq \frac{1}{f}\)

Thus, \(f^{-1}(f(x))=x\)

The inverse of function exists only when the function \(f\) is bijective. If the inverse of a function exists, then it is called an invertible function. The inverse of a function of a bijective function is unique.

Geometrically \(f^{-1}(x)\) is the image of \(f(x)\) concerning a line \(y=x\). In other words, \(f^{-1}(x)\) is symmetrical concerning the line \(y=x\).

A function \(f(x)\) is said to be involution if for all \(x\) for which \(f(x)\) and \(f(f(x))=x\)

If \(f\) is an invertible function, then \(\left(f^{-1}\right)^{-1}=f\)

If \(f: A \rightarrow B\) be a one-one function, then \(f^{-1} o f=I_{A}\) and \(f o f^{-1}=I_{B}\), where, \(I_{A}\) and \(I_{B}\) are the identity functions of the set \(A\) and \(B\) respectively.

Let \(f: A \rightarrow B, g: B \rightarrow C\) be two invertible functions, then \(g o f\) is also invertible with \((g o f)^{-1}=\left(f^{-1} o g^{-1}\right)\)

Learn Important Trigonometry Formulas

Steps to Find the Inverse of a Function

1. We need to check whether the given function is bijective or not
2. If the given function is bijective, then the inverse exists, other-wise not.
3. Find out \(x\) in terms of \(y\)
4. Substitute \(y\) by \(x\), then we get the inverse of \(f\), i.e., \(f^{-1}(x)\)

Inverse Trigonometric Functions and Graphs

We can define the trigonometric functions as below:

(i) sin: \(R \rightarrow[-1,1]\)
(ii) cos: \(R \rightarrow[-1,1]\)
(iii) tan: \(R – \left\{ {x:x = (2n + 1)\frac{\pi }{2},n \in Z} \right\} \to R\)
(iv) cot: \(R – \left\{ {x:x = n\pi ,\,n \in Z} \right\} \to R\)
(v) sec: \(R – \left\{ {x:x = (2n + 1)\frac{\pi }{2},n \in Z}  \right\} \to R – ( – 1,1)\)
(vi) cosec: \(R – \left\{ {x:x = n\pi ,\,n \in Z} \right\} \to R – \left( { – 1,\,1} \right)\)

The domain of the sine function is the set of all real numbers, and the range is the closed interval \([-1,1]\). Thus the graph of \(f(x)=\sin x\) is shown below.

From the graph, we can say that it will be one-one and onto only when we considered it in some particular intervals like \(\left[-\frac{3 \pi}{2},-\frac{\pi}{2}\right],\left[-\frac{\pi}{2}, \frac{\pi}{2}\right],\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]\) and so on. If we consider the whole function, then it is not one-one as well as onto.

Also, when we think of the inverse function, then the domain and range are interchanged. So the graph of this function is shown below.

As a whole, the inverse of this function does not exist. Its inverse exists only when we restrict its range. So the intervals are \(\left[-\frac{3 \pi}{2},-\frac{\pi}{2}\right],\left[-\frac{\pi}{2}, \frac{\pi}{2}\right],\left[\begin{array}{l} \left.\frac{\pi}{2}, \frac{3 \pi}{2}\right]\end{array}\right]\) and so on.

In general, we consider it as \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

So, sine inverse is defined as \(\sin ^{-1}:[-1,1] \rightarrow\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
The domain of the cosine function is the set of all real numbers, and the range is the closed interval \([-1,1]\). Thus the graph of \(f(x)=\cos x\) is shown below.

From the graph, we can say that it will be one-one and onto only when we considered it in some particular intervals like \((-\pi, 0),(0, \pi),(\pi, 2 \pi)\) and so on. If we consider the whole function, then it is not one-one as well as onto.

The graph of the function \(y=\cos ^{-1} x\) can be drawn in the same way as discussed the graph of \(y=\sin ^{-1} x\). The graph of \(\cos ^{-1} x\) is shown below.

We know that, \(\operatorname{cosec} x=\frac{1}{\sin x}\), the domain and range of cosecant function is \(R – \left\{ {x:x = n\pi ,\,n \in Z} \right\}\) and \(R-(-1,1)\) respectively. It means that \(y=\operatorname{cosec} x\) assumes all real values except \(-1<y<1\) and is not defined for integral multiple of \(\pi\). If we restrict the domain of the cosecant function to \(\left[ { – \frac{\pi }{2},\frac{\pi }{2}} \right] – \left\{ 0 \right\},\) then it is one-one and onto with its range as the set \(R-(-1,1)\).

The cosecant function which is restricted to any of the intervals \(\left[ { – \frac{{3\pi }}{2},\frac{{ – \pi }}{2}} \right] – \left\{ { – \pi } \right\},\,\left[ { – \frac{\pi }{2},\,\frac{\pi }{2}} \right] – \left\{ 0 \right\},\,\left[ {\frac{\pi }{2},\,\frac{{3\pi }}{2}} \right] – \left\{ \pi \right\}\) etc. is bijective, and its range is the set of all real numbers \(R-(-1,1)\).

The graph of \(y=\operatorname{cosec}^{-1} x\) is given below.

Also, \(\sec x=\frac{1}{\cos x}\), the domain of \(y=\sec x\) is \(R – \left\{ {x:x = \left( {2n + 1} \right)\frac{\pi }{2},\,n \in Z} \right\}\) and the range is \(R-(-1,1)\). This means the secant function assumes all real values except \(-1<y<1\) and is not defined for odd multiples of \(\frac{\pi}{2}\).

The secant function which is restricted to any of the intervals \(\left[ { – \pi ,\,0} \right] – \left\{ {\frac{{ – \pi }}{2}} \right\},\left[ {0,\,\pi } \right] – \left\{ {\frac{\pi }{2}} \right\},\left\{ {\pi ,\,2\pi } \right\} – \left\{ {\frac{{3\pi }}{2}} \right\}\) etc. is bijective, and its range is the set of all real numbers \(R-(-1,1)\).

Therefore, \(\sec ^{-1}\) can be defined as a function whose domain is \(R-(-1,1)\) and range could be any of the intervals \([ – \pi ,0] – \left\{ {\frac{{ – \pi }}{2}} \right\},[0,\pi ] – \left\{ {\frac{\pi }{2}} \right\},[\pi ,2\pi ] – \left\{ {\frac{{3\pi }}{2}} \right\}\) and so on.

The graph of \(y=\sec ^{-1} x\) are given below.

We know that the domain of the tan function is the set \(R – \left\{ {x:x = (2n + 1)\frac{\pi }{2},n \in Z} \right\}\) and the range is \(R\). This means that tan is not defined for odd multiples of \(\frac{\pi}{2}\). If we restrict the domain of a tangent function to \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), then it is one-one and onto with its range \(R\).

Tangent function restricted to any of the intervals \(\left(-\frac{3 \pi}{2}, \frac{-\pi}{2}\right),\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\) etc. is bijective, and its range is \(R\).

Therefore, \(\tan ^{-1}\) can be defined as a function whose domain is \(R\) and range could be any of the intervals \(\left(-\frac{3 \pi}{2}, \frac{-\pi}{2}\right),\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\) and so on.

The graph of \(y=\tan ^{-1} x\) is given below:

The domain of the cot function is the set \(R – \left\{ {x:x = n\pi ,\,n \in Z} \right\}\) and the range is \(R\). This means that the cotangent function is not defined for the integral multiple of \(\pi\). If we restrict the domain of the cotangent function to \((0, \pi)\), then it is bijective with its range as \(R\). If cotangent is restricted to any of the intervals \((-\pi, 0),(0, \pi),(\pi, 2 \pi)\) etc., is bijective and its range is \(R\).

Therefore, \(\cot ^{-1}\) can be defined as a function whose domain is \(R\) and range could be any of the intervals \((-\pi, 0),(0, \pi),(\pi, 2 \pi)\) and so on.

The graph of \(y=\cot ^{-1} x\) is given below.

Domain and Range of Inverse Trigonometric Functions

The below table gives the inverse trigonometric functions along with their domains and ranges.

Formulas of Inverse Trigonometric Functions

In this section, we will learn about some important properties of inverse trigonometric functions.

(i) \(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}, \forall x \in[-1,1]\)

(ii) \(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}, \forall x \in R\)

(iii) \(\operatorname{cosec}^{-1} x+\sec ^{-1} x=\frac{\pi}{2}, \forall x \in R-(-1,1)\)

(iv) \({\sin ^{ – 1}}x = {\rm{cose}}{{\rm{c}}^{ – 1}}\left( {\frac{1}{x}} \right),\,x \in \left[ { – 1,\,1} \right] – \left\{ 0 \right\}\)

(v) \(\operatorname{cosec}^{-1} x=\sin ^{-1}\left(\frac{1}{x}\right),|x| \geq 1\)

(vi) \({\cos ^{ – 1}}x = {\rm{se}}{{\rm{c}}^{ – 1}}\left( {\frac{1}{x}} \right),\,x \in \left[ { – 1,\,1} \right] – \left\{ 0 \right\}\)

(vii) \(\sec ^{-1} x=\cos ^{-1}\left(\frac{1}{x}\right),|x| \geq 1\)

(viii) \(\tan ^{-1} x=\cot ^{-1}\left(\frac{1}{x}\right): x>0\)

(ix) \(\cot ^{-1} x=\tan ^{-1}\left(\frac{1}{x}\right): x>0\)

(x) \(\sin ^{-1}(-x)=-\sin ^{-1} x, x \in[-1,1]\)

(xi) \(\tan ^{-1}(-x)=-\tan ^{-1} x, x \in R\)

(xii) \(\operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1} x,|x| \geq 1\)

(xii) \(\cos ^{-1}(-x)=\pi-\cos ^{-1} x, x \in[-1,1]\)

(xiii) \(\sec ^{-1}(-x)=\pi-\sec ^{-1} x,|x| \geq 1\)

(xiv) \(\cot ^{-1}(-x)=\pi-\cot ^{-1} x, x \in R\)

(xv) \(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}, x y<1\)

(xvi) \(\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}, x y>-1\)

(xvii) \(2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}, \quad|x| \leq 1\)

(xviii) \(2 \tan ^{-1} x=\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}, x \geq 0\)

(xix) \(2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}},-1<x<1\)

Derivatives of Inverse Trigonometric Functions

Below are some derivatives of inverse trigonometric functions:

(i) \(\frac{d}{{dx}}\left( {{{\sin }^{ – 1}}x} \right) = \frac{1}{{\sqrt {1 – {x^2}} }}\)

(ii) \(\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}}\)

(iii) \(\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}\)

(iv) \(\frac{d}{d x}\left(\cot ^{-1} x\right)=\frac{-1}{1+x^{2}}\)

(v) \(\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{x \sqrt{x^{2}-1}}\)

(vi) \(\frac{d}{d x}\left(\operatorname{cosec}^{-1} x\right)=\frac{-1}{x \sqrt{x^{2}-1}}\)

Example: If \(y=\tan ^{-1}\left(\frac{1-\cos x}{\sin x}\right)\), find \(\frac{d y}{d x}\).

Solution: Given: \(y=\tan ^{-1}\left(\frac{1-\cos x}{\sin x}\right)\)
We know that, \(1-\cos x=2 \sin ^{2} \frac{x}{2}\) and \(\sin \,x = 2\,\sin \frac{x}{2}\cos \frac{x}{2}\)

So, \(y = {\tan ^{ – 1}}\left( {\frac{{2\,{{\sin }^2}\frac{x}{2}}}{{2\,\sin \frac{x}{2}\cos \frac{x}{2}}}} \right)\)
\(y = {\tan ^{ – 1}}\left( {\frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}}} \right)\)
\(\Rightarrow y=\tan ^{-1}\left(\tan \frac{x}{2}\right)\)
\(\Rightarrow y=\frac{x}{2}\)

Now, differentiating with respect to \(x\), we get

\(\frac{d y}{d x}=\frac{1}{2}\)

Integration of Inverse Trigonometric Functions

Below are some of the important formulas of inverse trigonometric functions in the integration.

(i) \(\int \frac{d x}{\sqrt{1-x^{2}}}=\sin ^{-1} x+c\)

(ii) \(\int \frac{-d x}{\sqrt{1-x^{2}}}=\cos ^{-1} x+c\)

(iii) \(\int \frac{d x}{1+x^{2}}=\tan ^{-1} x+c\)

(iv) \(\int \frac{-d x}{1+x^{2}}=\cot ^{-1} x+c\)

(v) \(\int \frac{d x}{x \sqrt{x^{2}-1}}=\sec ^{-1} x+c\)

(vi) \(\int \frac{-d x}{x \sqrt{x^{2}-1}}=\operatorname{cosec}^{-1} x+c\)

Example: Find \(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x\)

Solution: First we find the integral of \(\int \frac{x}{\sqrt{1-x^{2}}} d x\)
Put \(t=1-x^{2}\)
\(\Rightarrow d t=-2 x d x\)
So, \(\int \frac{x}{\sqrt{1-x^{2}}} d x=-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\sqrt{t}=-\sqrt{1-x^{2}}\)
Hence, \(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x=\sin ^{-1} x\left(-\sqrt{1-x^{2}}\right)-\int \frac{1}{\sqrt{1-x^{2}}}\left(-\sqrt{1-x^{2}}\right) d x\)
\(=-\sqrt{1-x^{2}} \sin ^{-1} x+x+C\)

Therefore, \(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x=x-\sqrt{1-x^{2}} \sin ^{-1} x+C\)

Solved Examples– Inverse Trigonometric Functions

Q.1. Prove that \(\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}\)
Ans: \(\mathrm{LHS}: \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)\)
Put \(x=\sin \theta \Rightarrow \theta=\sin ^{-1} x\)
\(\Rightarrow \sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right)\)
We know that, \(1-\sin ^{2} \theta=\cos ^{2} \theta\)
\(\Rightarrow \sin ^{-1}\left(2 \sin \theta \sqrt{\cos ^{2} \theta}\right)\)
\(\Rightarrow \sin ^{-1}(2 \sin \theta \cos \theta)\)
Also, \(2 \sin \theta \cos \theta=\sin 2 \theta\)
\(\Rightarrow \sin ^{-1}(\sin 2 \theta)\)
\(\Rightarrow 2 \theta\)
\(\Rightarrow 2 \sin ^{-1} x=\) RHS
Hence, \(\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x\)

Q.2. Prove that \(3 \cos ^{-1} x=\cos ^{-1}\left(4 x^{3}-3 x\right)\)
Ans: RHS: \(\cos ^{-1}\left(4 x^{3}-3 x\right)\)
Put \(x=\cos \theta \Rightarrow \theta=\cos ^{-1} x\)
\(\Rightarrow \cos ^{-1}\left(4 \cos ^{3} \theta-3 \cos \theta\right)\)
We know that, \(4 \cos ^{3} \theta-3 \cos \theta=\cos 3 \theta\)
\(\Rightarrow \cos ^{-1}(\cos 3 \theta)\)
\(\Rightarrow 3 \theta\)
From (i) \(\theta=\cos ^{-1} x\)
So, \(3 \theta=3 \cos ^{-1} x=\) LHS

Q.3. Prove that \(2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}\)
Ans: We know that, \(2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\)
Therefore, \(2 \tan ^{-1} \frac{1}{2}=\tan ^{-1} \frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\)
\(\Rightarrow 2 \tan ^{-1} \frac{1}{2}=\tan ^{-1} \frac{1}{\frac{4-1}{4}}\)
\(\Rightarrow 2 \tan ^{-1} \frac{1}{2}=\tan ^{-1} \frac{4}{3}\)
Substituting \(2 \tan ^{-1} \frac{1}{2}=\tan ^{-1} \frac{4}{3}\) in the LHS of the given equation, we get
\(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{7}\)
Now, we need to prove \(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}\)
Also, \(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\)
So, \(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{\frac{4}{3}+\frac{1}{7}}{1-\left(\frac{4}{3} \times \frac{1}{7}\right)}\)
\(=\tan ^{-1} \frac{\frac{28+3}{21}}{1-\frac{4}{21}}\)
\(=\tan ^{-1} \frac{\frac{31}{21}}{\frac{21-4}{21}}\)
\(=\tan ^{-1} \frac{31}{17}\)
So, \(2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}\)

Q.4. Write \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), 0<x<\pi\) in the simplest form.
Ans: Given: \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), 0<x<\pi\)
We know that, \(1-\cos x=2 \sin ^{2} \frac{x}{2}\) and \(1+\cos x=2 \cos ^{2} \frac{x}{2}\)
\(\Rightarrow \tan ^{-1}\left(\sqrt{\frac{2 \sin ^{2 \frac{x}{2}}}{2 \cos ^{2} \frac{x}{2}}}\right)\)
\(\Rightarrow \tan ^{-1}\left(\sqrt{\tan ^{2} \frac{x}{2}}\right)\)
\(\Rightarrow \tan ^{-1}\left(\tan \frac{x}{2}\right)\)
\(=\frac{x}{2}\)
Therefore, the simplest form of \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\) is \(\frac{x}{2}\)

Q.5. Find the value of \(\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]\)
Ans: We know that the principal value of \(\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}\)
\(\Rightarrow \tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]=\tan ^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]\)
\(=\tan ^{-1}\left[2 \cos \left(\frac{\pi}{3}\right)\right]\)
We know that, \(\cos \left(\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Rightarrow \tan ^{-1}\left[2 \cos \left(\frac{\pi}{3}\right)\right]=\tan ^{-1}\left[2 \times \frac{1}{2}\right]\)
\(=\tan ^{-1} 1\)
\(=\frac{\pi}{4}\)
Therefore, \(\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]=\frac{\pi}{4}\)

Summary

In this article, we have learnt the meaning of inverse of functions, steps to find the inverse of functions, inverse trigonometric functions, graphs of inverse trigonometric functions and properties of trigonometric functions.

We have also learnt the formulas of derivatives and the integration of inverse trigonometric functions. Also, we solved some example problems based on the properties of inverse trigonometric functions. 

Applications of Trigonometry in Our Daily Life

Frequently Asked Questions

Q.1. What is inverse trigonometric functions?
Ans: The method to find the inverse functions of the trigonometric functions is known as inverse trigonometric functions.

Q.2. How to find the principal value of inverse trigonometric functions?
Ans: Let us consider an example to find the principal value of \(\sin ^{-1} \frac{\sqrt{3}}{2}\).
Principal value of \(\sin ^{-1} \frac{\sqrt{3}}{2}\) means the angle \(\theta\) where \(\sin \theta=\frac{\sqrt{3}}{2}\), we know that the range of \(\sin ^{-1}\) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and \(\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\). Clearly \(\frac{\pi}{3}\) lies between \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). So, the principal value of \(\sin ^{-1} \frac{\sqrt{3}}{2}\) is \(\frac{\pi}{3}\).
The principal value branch of the inverse trigonometric functions are as below:

Q.3. How to differentiate inverse trigonometric functions?
Ans: We have some basic derivatives of inverse trigonometric functions. By using these derivatives or formulas, we can differentiate the given inverse trigonometric functions. They are:
\(\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}\)
\(\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}}\)
\(\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}\)
\(\frac{d}{d x}\left(\cot ^{-1} x\right)=\frac{-1}{1+x^{2}}\)
\(\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{x \sqrt{x^{2}-1}}\)
\(\frac{d}{d x}\left(\operatorname{cosec}^{-1} x\right)=\frac{-1}{x \sqrt{x^{2}-1}}\)

Q.4. How to find the domain of inverse trigonometric functions?
Ans: The domain of the inverse trigonometric function is the range of the original trigonometric function. By using the table below, we can find the range and domain of the inverse trigonometric functions.

Q.5. Are inverse trigonometric functions always differentiable?
Ans: In their respective domain, all the trigonometric functions and inverse trigonometric functions are differentiable.

We hope this detailed article on inverse trigonometric functions helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!