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Inverse Trigonometric Functions class 12 Maths solutions are provided here. These NCERT solutions are created by expert team at Careers360 considering the demand of class 12 students, CBSE latest syllabus and pattern 2023-23, and comprehensively covering step by step all the concepts. Therefore these are great resource for students to command the concepts and in-depth understanding of the concepts. These inverse trigonometric functions class 12 NCERT solutions is one of the most important chapters in the NCERT Class 12 Maths book. These class 12 Maths chapter 2 notes will be beneficial for board exams as well as for other entrance exams such as JEE.
Concepts of maths chapter 2 class 12 are useful in other concepts like differential calculus and Integral calculus therefore class 12 inverse trigonometry becomes very important for students as it is base. Inverse trigonometric functions of NCERT class 12 chapter 2 also include exercises that are important as they help in practicing the problems and in depth understanding of class 12 chapter 2 inverse trigonometric functions of ncert.
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>> Basic Concepts: Inverse trigonometric functions map real numbers back to angles.
The inverse of the sine function: sin^{-1}(x) or arcsin(x) is defined on [-1, 1].
Properties of Inverse Trigonometric Functions:
Function | Domain | Range |
y = sin^{-1}(x) | [-1, 1] | [-π/2, π/2] |
y = cos^{-1}(x) | [-1, 1] | [0, π] |
y = cosec^{-1}(x) | R - (-1, 1) | [-π/2, π/2] - {0} |
y = sec^{-1}(x) | R - (-1, 1) | [0, π] - {π/2, π/2} |
y = tan^{-1}(x) | R | (-π/2, π/2) |
y = cot^{-1}(x) | R | (0, π) |
Self-Adjusting Trigonometric Property:
sin(sin^{-1}(x)) = x
sin^{-1}(sin(x)) = x
cos(cos^{-1}(x)) = x
cos^{-1}(cos(x)) = x
tan(tan^{-1}(x)) = x
tan^{-1}(tan(x)) = x
sec(sec^{-1}(x)) = x
sec^{-1}(sec(x)) = x
cosec^{-1}(cosec(x)) = x
cosec(cosec^{-1}(x)) = x
cot^{-1}(cot(x)) = x
cot(cot^{-1}(x)) = x
Reciprocal Relations:
sin^{-1}(1/x) = cosec^{-1}(x), x ≥ 1 or x ≤ -1
cos^{-1}(1/x) = sec^{-1}(x), x ≥ 1 or x ≤ -1
tan^{-1}(1/x) = cot^{-1}(x), x > 0
Even and Odd Functions:
sin^{-1}(-x) = -sin^{-1}(x), x ∈ [-1, 1]
tan^{-1}(-x) = -tan^{-1}(x), x ∈ R
cosec^{-1}(-x) = -cosec^{-1}(x), |x| ≥ 1
cos^{-1}(-x) = π - cos^{-1}(x), x ∈ [-1, 1]
sec^{-1}(-x) = π - sec^{-1}(x), |x| ≥ 1
cot^{-1}(-x) = π - cot^{-1}(x), x ∈ R
Complementary Relations:
sin^{-1}(x) + cos^{-1}(x) = π/2
tan^{-1}(x) + cot^{-1}(x) = π/2
cosec^{-1}(x) + sec^{-1}(x) = π/2
Sum and Difference Formulas:
tan^{-1}(x) + tan^{-1}(y) = tan^{-1}((x+y)/(1-xy))
tan^{-1}(x) - tan^{-1}(y) = tan^{-1}((x-y)/(1+xy))
sin^{-1}(x) + sin^{-1}(y) = sin^{-1}[x√(1-y^{2})+y√(1-x^{2})]
sin^{-1}(x) - sin^{-1}(y) = sin^{-1}[x√(1-y^{2})-y√(1-x^{2})]
cos^{-1}(x) + cos^{-1}(y) = cos^{-1}[xy-√(1-x^{2})√(1-y^{2})]
cos^{-1}(x) - cos^{-1}(y) = cos^{-1}[xy+√(1-x^{2})√(1-y^{2})]
cot^{-1}(x) + cot^{-1}(y) = cot^{-1}((xy-1)/(x+y))
cot^{-1}(x) - cot^{-1}(y) = cot^{-1}((xy+1)/(y-x))
Double Angle Formula:
2 tan^{-1}(x) = sin^{-1}(2x/(1+x^{2}))
2 tan^{-1}(x) = cos^{-1}((1-x^{2})/(1+x^{2}))
2 tan^{-1}(x) = tan^{-1}(2x/(1-x^{2}))
2 sin^{-1}(x) = sin^{-1}(2x√(1+x^{2}))
2 cos^{-1}(x) = sin^{-1}(2x√(1-x^{2}))
Conversion Properties:
sin^{-1}(x) = cos^{-1}(√(1-x^{2})) = tan^{-1}(x/√(1-x^{2})) = cot^{-1}(√(1-x^{2})/x)
cos^{-1}(x) = sin^{-1}(√(1-x^{2})) = tan^{-1}(√(1-x^{2})/x) = cot^{-1}(x/√(1-x^{2}))
tan^{-1}(x) = sin^{-1}(x/√(1-x^{2})) = cos^{-1}(x/√(1+x^{2})) = sec^{-1}(√(1+x^{2})) = cosec^{-1}(√(1+x^{2})/x)
Free download Inverse Trigonometric Functions Class 12 NCERT solutions for CBSE Exam.
Inverse Trigonometric Functions Class 12 Solutions: Exercise 2.1
Question:1 Find the principal values of the following :
Answer:
Let
We know, principle value range of is
The principal value of is
Question:2 Find the principal values of the following:
Answer:
So, let us assume that then,
Taking inverse both sides we get;
, or
and as we know that the principal values of is from [0, ],
Hence when x = .
Therefore, the principal value for is .
Question:3 Find the principal values of the following:
Answer:
Let us assume that , then we have;
, or
.
And we know the range of principal values is
Therefore the principal value of is .
Question:4 Find the principal values of the following:
Answer:
Let us assume that , then we have;
or
and as we know that the principal value of is .
Hence the only principal value of when .
Question:5 Find the principal values of the following:
Answer:
Let us assume that then,
Easily we have; or we can write it as:
as we know that the range of the principal values of is .
Hence lies in the range it is a principal solution.
Question:6 Find the principal values of the following :
Answer:
Given so we can assume it to be equal to 'z';
,
or
And as we know the range of principal values of from .
As only one value z = lies hence we have only one principal value that is .
Question:7 Find the principal values of the following :
Answer:
Let us assume that then,
we can also write it as; .
Or and the principal values lies between .
Hence we get only one principal value of i.e., .
Question:8 Find the principal values of the following:
Answer:
Let us assume that , then we can write in other way,
or
.
Hence when we have .
and the range of principal values of lies in .
Then the principal value of is
Question:9 Find the principal values of the following:
Answer:
Let us assume ;
Then we have
or
,
.
And we know the range of principal values of is .
So, the only principal value which satisfies is .
Question:10 Find the principal values of the following:
Answer:
Let us assume the value of , then
we have or
.
and the range of the principal values of lies between .
hence the principal value of is .
Question:11 Find the values of the following:
Answer:
To find the values first we declare each term to some constant ;
, So we have ;
or
Therefore,
So, we have
.
Therefore ,
,
So we have;
or
Therefore
Hence we can calculate the sum:
.
Question:12 Find the values of the following:
Answer:
Here we have
let us assume that the value of
;
then we have to find out the value of x +2y.
Calculation of x :
,
Hence .
Calculation of y :
.
Hence .
The required sum will be = .
Question:13 If then
Answer:
Given if then,
As we know that the can take values between
Therefore, .
Hence answer choice (B) is correct.
Question:14 is equal to
Answer:
Let us assume the values of be 'x' and be 'y'.
Then we have;
or or or
.
and or
or
also, the ranges of the principal values of and are . and
respectively.
we have then;
Inverse Trigonometric Functions Class 12 Solutions: Exercise 2.2
Question:1 Prove the following:
Answer:
Given to prove:
where, .
Take or
Take R.H.S value
=
=
=
= = L.H.S
Question:2 Prove the following:
Answer:
Given to prove .
Take or ;
Then we have;
R.H.S.
=
=
=
= = L.H.S
Hence Proved.
Question:4 Prove the following:
Answer:
Given to prove
Then taking L.H.S.
We have
= R.H.S.
Hence proved.
Question:5 Write the following functions in the simplest form:
Answer:
We have
Take
is the simplified form.
Question:7 Write the following functions in the simplest form:
Answer:
Given that
We have in inside the root the term :
Put and ,
Then we have,
Hence the simplest form is
Question:8 Write the following functions in the simplest form:
Answer:
Given where
So,
Taking common from numerator and denominator.
We get:
= as,
= is the simplest form.
Question:9 Write the following functions in the simplest form:
Answer:
Given that
Take or
and putting it in the equation above;
is the simplest form.
Question:10 Write the following functions in the simplest form:
Answer:
Given
Here we can take
So,
will become;
and as ;
hence the simplest form is .
Question:11 Find the values of each of the following:
Answer:
Given equation:
So, solving the inner bracket first, we take the value of
Then we have,
Therefore, we can write .
.
Question:12 Find the values of each of the following:
Answer:
We have to find the value of
As we know so,
Equation reduces to .
Question:13 Find the values of each of the following: and
Answer:
Taking the value or and or then we have,
= ,
=
Then,
Ans.
Question:14 If , then find the value of .
Answer:
As we know the identity;
. it will just hit you by practice to apply this.
So, or ,
we can then write ,
putting in above equation we get;
=
Ans.
Question:15 If , then find the value of .
Answer:
Using the identity ,
We can find the value of x;
So,
on applying,
=
= or ,
Hence, the possible values of x are .
Question:16 Find the values of each of the expressions in Exercises 16 to 18.
Answer:
Given ;
We know that
If the value of x belongs to then we get the principal values of .
Here,
We can write is as:
=
= where
Question:17 Find the values of each of the expressions in Exercises 16 to 18.
Answer:
As we know
If which is the principal value range of .
So, as in ;
Hence we can write as :
=
Where
and
Question:18 Find the values of each of the expressions in Exercises 16 to 18.
Answer:
Given that
we can take ,
then
or
We have similarly;
Therefore we can write
from
Question:19 is equal to
Answer:
As we know that if and is principal value range of .
In this case ,
hence we have then,
Hence the correct answer is (B).
Question:20 is equal to
Answer:
Solving the inner bracket of ;
or
Take then,
and we know the range of principal value of
Therefore we have .
Hence,
Hence the correct answer is D.
Question:21 is equal to
Answer:
We have ;
finding the value of :
Assume then,
and the range of the principal value of is .
Hence, principal value is
Therefore
and
so, we have now,
or,
Hence the answer is option (B).
NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions: Miscellaneous Exercise
Question:1 Find the value of the following:
Answer:
If then , which is principal value of .
So, we have
Therefore we have,
.
Question:2 Find the value of the following:
Answer:
We have given ;
so, as we know
So, here we have .
Therefore we can write as:
.
Question:3 Prove that
Answer:
To prove: ;
Assume that
then we have .
or
Therefore we have
Now,
We can write L.H.S as
as we know
L.H.S = R.H.S
Question:4 Prove that
Answer
Taking
then,
Therefore we have-
.............(1).
,
Then,
.............(2).
So, we have now,
L.H.S.
using equations (1) and (2) we get,
= R.H.S.
Question:5 Prove that
Answer:
Take and and
then we have,
Then we can write it as:
or
...............(1)
Now,
So, ...................(2)
Also we have similarly;
Then,
...........................(3)
Now, we have
L.H.S
so, using (1) and (2) we get,
or we can write it as;
= R.H.S.
Hence proved.
Question:6 Prove that
Answer:
Converting all terms in tan form;
Let , and .
now, converting all the terms:
or
We can write it in tan form as:
.
or ................(1)
or
We can write it in tan form as:
or ......................(2)
Similarly, for ;
we have .............(3)
Using (1) and (2) we have L.H.S
On applying
We have,
...........[Using (3)]
=R.H.S.
Hence proved.
Question:7 Prove that
Answer:
Taking R.H.S;
We have
Converting sin and cos terms in tan forms:
Let and
now, we have or
............(1)
Now,
................(2)
Now, Using (1) and (2) we get,
R.H.S.
as we know
so,
equal to L.H.S
Hence proved.
Question:8 Prove that
Answer:
Applying the formlua:
on two parts.
we will have,
Hence it s equal to R.H.S
Proved.
Question:9 Prove that
Answer:
By observing the square root we will first put
.
Then,
we have
or, R.H.S.
.
L.H.S.
hence L.H.S. = R.H.S proved.
Question:10 Prove that
Answer:
Given that
By observing we can rationalize the fraction
We get then,
Therefore we can write it as;
As L.H.S. = R.H.S.
Hence proved.
Question:11 Prove that
Answer:
By using the Hint we will put ;
we get then,
dividing numerator and denominator by ,
we get,
using the formula
As L.H.S = R.H.S
Hence proved
Question:12 Prove that
Answer:
We have to solve the given equation:
Take as common in L.H.S,
or from
Now, assume,
Then,
Therefore we have now,
So we have L.H.S then
That is equal to R.H.S.
Hence proved.
Question:13 Solve the following equations:
Answer:
Given equation ;
Using the formula:
We can write
So, we can equate;
that implies that .
or or
Hence we have solution .
Question:14 Solve the following equations:
Answer:
Given equation is
:
L.H.S can be written as;
Using the formula
So, we have
Hence the value of .
Question:16 then is equal to
Answer:
Given the equation:
we can migrate the term to the R.H.S.
then we have;
or ............................(1)
from
Take or .
So, we conclude that;
Therefore we can put the value of in equation (1) we get,
Putting x= sin y , in the above equation; we have then,
So, we have the solution;
Therefore we have .
When we have , we can see that :
So, it is not equal to the R.H.S.
Thus we have only one solution which is x = 0
Hence the correct answer is (C).
If you are looking for inverse trigonometric functions class 12 exercises then these are listed below.
Inverse Trigonometric Functions Class 12 Exercise 2.1
Inverse Trigonometric Functions Class 12 Exercise 2.2
Inverse Trigonometric Functions Class 12 Miscellaneous Exercise
NCERT Solutions for Class 12 Maths Chapter 2 Introduction
In class 11 Maths you have already learnt about trigonometric functions. It won't take much effort to command on inverse trigonometric functions if you have good knowledge of trigonometric functions. You just need to practice NCERT questions including examples and miscellaneous exercises given in inverse trigonometric functions class 12. You may find some difficulties in solving the problems, so you can take the help of inverse trigonometry class 12 solutions maths chapter 2 notes.
There are important applications of ITF in geometry, navigation, science, and engineering. Also, inverse trigonometric functions play an important role in the calculus part of mathematics to define many integrals. Many students have a misconception in class 12 maths ch 2 question answer function like .But the inverse function is not the same as for example .
For inverse to exist, the function must be one-one and onto but trigonometric functions are neither one-one and onto over their domain and natural ranges. So, to ensure the existence of their inverse we restrict domains and ranges of trigonometric functions. And this range is known as principal value.
The important topics covered in the class 12 maths ch 2 question answer are
Basic concepts
This chapter of class 12 NCERT includes concepts of domain and range of trigonometric function in which their inverse exists. the chapter comprehensively discusses graphical representation, the elementary property of inverse trigonometry function.
ch 2 maths class 12 discusses the basic concepts of the inverse of the sine function, cosine function, secant function, cosecant function, tangent function, and cotangent function. Also, we will be familiar with their range and domain, concepts related to one-one, and onto function. Problems related to these concepts are in detail in NCERT solutions for class 12 maths chapter 2.
Properties of inverse trigonometry function
Maths class 12 chapter 2 includes important property of inverse trigonometry function that might be valid in principal value branch and corresponding inverse trigonometric function. To get a good hold on these concepts you can solve the NCERT solution for class 12 maths chapter 2.
Topics mentioned in class 12 ncert are very important to obtain decent marks in board exams and students are recommended to proceed through all the concepts cited in the topics. Questions from all the above topics are covered in the NCERT solutions for class 12 maths chapter 2.
NCERT Exemplar Class 12 Solutions
Chapter 1 | NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions |
Chapter 2 | NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions |
Chapter 3 | NCERT solutions for class 12 maths chapter 3 Matrices |
Chapter 4 | NCERT solutions for class 12 maths chapter 4 Determinants |
Chapter 5 | NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability |
Chapter 6 | NCERT solutions for class 12 maths chapter 6 Application of Derivatives |
Chapter 7 | NCERT solutions for class 12 maths chapter 7 Integrals |
Chapter 8 | NCERT solutions for class 12 maths chapter 8 Application of Integrals |
Chapter 9 | NCERT solutions for class 12 maths chapter 9 Differential Equations |
Chapter 10 | NCERT solutions for class 12 maths chapter 10 Vector Algebra |
Chapter 11 | NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry |
Chapter 12 | NCERT solutions for class 12 maths chapter 12 Linear Programming |
Chapter 13 | NCERT solutions for class 12 maths chapter 13 Probability |
In class 11 maths you have already learnt about trigonometric functions. It won't take much effort to command inverse trigonometric functions if you have good knowledge of trigonometric functions.
For this, you just need to practice NCERT questions including examples and miscellaneous exercise.
You may find some difficulties in solving the problems, so you can take the help of these NCERT Class 12 Maths solutions chapter 2.
NCERT Solutions for Class 12 Maths Chapter 2 PDF Download will also be available soon to help you with the questions offline.
Inverse Trigonometric Functions Class 12 Solutions - Topics
2.1 Introduction
2.2 Basic Concepts
2.3 Properties of Inverse Trigonometric Functions
If you wish to perform well in the CBSE 12 board examination, NCERT Class 12 Maths solutions chapter 2 inverse trigonometric functions are very helpful but here are some tips to make command on Inverse Trigonometric Functions.
The inverse trigonometric function is inverse of the trigonometric function, so if you have a command on the trigonometric function then it will be easy for you to understand inverse trigonometric functions, use class 12 maths chapter 2 question answer pdf to learn easily.
Try to relate trigonometric functions formulas with inverse trigonometric functions formulas, so that memorizing the formulae becomes easier.
Before starting to solve an exercise, first solve the examples that are given in the NCERT class 12 maths textbook.
Also, try to solve every exercise including inverse trigonometric functions class 12 miscellaneous, and examples, miscellaneous examples on your own, if you are finding difficulties, you can take the help of Class 12 Maths Chapter 2 NCERT solutions.
If you have solved all NCERT then you can solve previous years paper CBSE board to get familiar with the pattern of the board exam question paper.
Also, check NCERT solutions for class 12 for other chapter and subjects.
Some basic concepts of inverse trigonometry and properties of inverse trigonometric functions are important topics in this chapter. these concepts are very important as they are used in solving or simplifying differential calculus, integral calculus, and other topics like differential equations therefore having a command on inverse trigonometry concepts becomes very important. students can refer to class 12 maths chapter 2 question answer for getting a good hold on the concepts.
NCERT class 12 maths ch 2 solutions provide students with a strong foundation in basic Maths topics. These exercise-wise class 12 maths ch 2 ncert solutions are crafted by experienced and knowledgeable subject experts at Carresr360. By using these resources, students can improve their focus and achieve better scores on their board exams. The ultimate goal is to enhance students' confidence and ability to efficiently solve difficult problems in a timely manner.
Relation and function and inverse trigonometry combined have 10 % weightage in the CBSE class 12 board examination. According to priority students can give time to chapters but if you want full marks then it demand practice and in depth understanding of concepts therefor ncert textbook are suggested to students.
Students consider Integration and it's application as the most difficult unit in CBSE class 12 maths. With regular practice, you will get conceptual clarity and will be able to have a strong grip on Integration. if you want then inverse trigonometric functions solutions are recommended to students for getting command and in depth understanding of concepts.
NCERT textbook is the best book for CBSE class 12 maths. Most of the questions in the CBSE class 12 board exam are directly asked from the NCERT textbook. All you need to do is practice of the questions given in the NCERT textbook that is best way to get command on concepts. if you have any difficulty then you can read solutions of ncert exercise also help you in getting hold on the chapter.
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