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Edited By Kuldeep Maurya | Updated on Jan 21, 2022 12:24 PM IST

You can find free RD Sharma class 12 chapter 3 Exercise 3.6 solution here. For Inverse Trigonometric Functions, all RD Sharma Book Solutions are listed here, exercise by exercise. RD Sharma Solutions can assist you with preparing for a variety of competitive exams in high school, graduate school, and undergraduate education. In addition, it has been proven that practicing questions with RD Sharma Class 12th Exercise 3.6 Solution improves math skills**.**

- Chapter 3 - Inverse Trigonometric Functions Ex 3.1
- Chapter 3 - Inverse Trigonometric Functions Ex 3.2
- Chapter 3 - Inverse Trigonometric Functions Ex 3.3
- Chapter 3 - Inverse Trigonometric Functions Ex 3.4
- Chapter 3 - Inverse Trigonometric Functions Ex 3.5
- Chapter 3 - Inverse Trigonometric Functions Ex 3.7
- Chapter 3 - Inverse Trigonometric Functions Ex 3.8
- Chapter 3 - Inverse Trigonometric Functions Ex 3.9
- Chapter 3 - Inverse Trigonometric Functions Ex 3.10
- Chapter 3 - Inverse Trigonometric Functions Ex 3.11
- Chapter 3 - Inverse Trigonometric Functions Ex 3.12
- Chapter 3 - Inverse Trigonometric Functions Ex 3.13
- Chapter 3 - Inverse Trigonometric Functions Ex 3.14
- Chapter 3 - Inverse Trigonometric Ex VSA

Inverse Trigonometric Function Exercise 3.6 Question 1 (i).

Thus,

Let

(From equation 1)

The principal value branch of the function

Hence the principal value of is

Inverse Trigonometric Function Exercise 3.6 Question 1 (ii).

Let

(From equation 1)

The principal value branch of the function is

Hence the principal value of is

Inverse Trigonometric Function Exercise 3.6 Question 1 (iii).

Let

(From equation 1)

The principal value branch of the function is

Hence the principal value of is

Inverse Trigonometric Function Exercise 3.6 Question 1 (iv).

**Answer:****Hint:** The function is defined as a function whose domain R and the principal value branch of the function is Thus

First we will convert into **Given: ****Solution:**

Let

(From equation 1)

The principal value branch of the function is

Hence the principal value of is

Inverse Trigonometric Function Exercise 3.6 Question 2.

Let us first solve for

is defined for

Domain of

Let us solve for

is defined for

Domain of

Now to find the domain of we take the intersection of both equations 1 and 2

(Domain of ) (Domain of )

Hence the domain of is

Inverse Trigonometric Function Exercise 3.6 Question 3 (i).

The principal value branch of the function

The principal value branch of the function

The principal value branch of the function cosec -1 is

We will find all the principal values between these intervals

Let us first solve for

Let

{from equation (ii)}

The principal value branch of

Therefore principal value of is

Let us solve for

Let

{from equation (iv)}

The principal value branch of

Therefore principal value of … (v)

Let us solve for

Let

{from equation (vi)}

The principal value branch of the function is

Therefore the principal value of … (vii)

Now from equation (i) –

Putting the value of from equations (iii), (v) and (vii) respectively.

Inverse Trigonometric Function Exercise 3.6 Question 3 (ii).

First we will find the principal value of

The principal value branch of function

The principal value branch of function

Let us first solve for

Let

[from equation (ii)]

The principal value branch of is

Principal value of is

Now from equation (i)

[from equation (iii)]

Let

[from v]

From equation (iv)

Inverse Trigonometric Function Exercise 3.6 Question 3 (iii).

The principal value branch of the function

The principal value branch of the function

Let us first solve for

Let

[from equation (ii)]

The principal value branch of the function

Therefore the principal value of is

Now let us solve for

Let

[from equation (iv) ]

The principal value branch of the is

Principal value of

Now, from equation (i)

From equation (iii) and (v)

Inverse Trigonometric Function Exercise 3.6 Question 3 (iv).

**Answer:****Hints:** First we will find principal values of **Given: ****Solution:**

Let

[from equation]

The range of principal value branch of is

Hence principal value of

Now let us solve for

Let

{from equation (iv)}

The principal value branch of is

Therefore principal value of

Now let us solve for

Let

The range of principal value branch of

The principal value of

Now from equation (i)

Putting the value of from equations (iii), (v) and (vii) respectively.

Class 12 RD Sharma Chapter 3 Exercise 3.6 solution has nine questions, including subparts, which cover questions regarding the principal value, domain, and range of inverse trigonometric functions, as well as several key qualities that can aid you in solving inverse trigonometric function problems. RD Sharma Class 12th Exercise 3.6, inverse trigonometric functions can be used to calculate the angle from any trigonometric ratio. All of these proportions have applications in various domains, including engineering, geometry, and physics.

Experts in the field create answers based on the students' comprehension abilities. Thus, it aids students in honing their logical reasoning abilities, which are critical for exam success. The inverse of the cotangent function is the sole topic of this practice. RD Sharma Class 12 solution Inverse Trigonometric Function Ex 3.6 PDF can be used as a study aid by students who desire to do well on the exam.

**Chapter-wise RD Sharma Class 12 Solutions**

- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable

1. Are the RD Sharma Class 12th Exercise 3.6 solutions up to current with the most recent syllabus?

RD Sharma solutions are the best option for exam preparation as they are accurate and updated to the latest version.

2. How do I get Class 12 RD Sharma chapter 3 exercise 3.6 solution?

It is extremely simple to obtain the RD Sharma class 12 solution chapter 3 exercise 3.6 solution for class 12 students. All you have to do is go to Career360's website and get the book's free ebook.

3. Is it affordable to use RD Sharma's book?

The RD Sharma Class 12th Exercise 3.6 material is available for free on Career360’s website. They require no additional charges and are meant to help students prepare well in exams.

Mar 22, 2023

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