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Edited By Kuldeep Maurya | Updated on Jan 21, 2022 12:33 PM IST

The CBSE Board students widely use the RD Sharma Solution books to clear their doubts in their homework and score good marks in their public examinations. However, every student cannot depend on a tutor to help them clear their doubts 24 x 7. Significantly, the students who find it difficult to cope with the Inverse Trigonometry portions can use the RD Sharma Class 12th exercise 3.8 books. ** **

- 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 3.6
- Chapter 3 - Inverse Trigonometric Functions Ex 3.7
- 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 trigromtery functions exercise 3.8 question 1 (I)

As we know that the value of

We have,

So here in the expression already present the form of

Inverse trigromtery functions exercise 3.8 question 1 (II)

As we know that the value of

We have,

Here in the place of there is .

So we convert into .

Let’s suppose that

Let, in, angle and right angle at B

[we will ignore the -ve sign because BC is a length and it can’t be -ve.]

So, we required and that is

Inverse trigromtery functions exercise 3.8 question 1 (III)

As we know that the value of

We have

Here in the place of we have .

So, we convert to .

Let’s suppose that,

Let, in , angle and right angle at B.

[we will ignore the -ve sign because AC is a length and it can’t be -ve.]

[since, ]

Inverse trigonometric functions exercise 3.8 question 1 (iv)

As we know that the value of

We have

So here in place of there is .

So, we convert into

Let’s we suppose that,

Let, in , angle and right angle at B.

[we will ignore the -ve sign because BC is a length and it can’t be -ve]

Inverse trigonometric functions exercise 3.8 question 1 (v)

As we know that the value of

We have

Solution:

In place of here is

So, we convert into .

Let’s suppose that,

Let, in , angle and right angle at B.

[we will ignore the -ve sign because BC is a length and it can’t be -ve]

Inverse trigonometric functions exercise 3.8 question 1 (vi)

As we know that the range of

We have

In place of there is

So, we convert into .

Let’s suppose that,

Let, in , angle and right angle at B.

[we will ignore the -ve sign because AB is a length and it can’t be -ve]

Inverse Trigonometric functions exercise 3.8 question 1 (vii)

As we know that the value of

We have

In place of here is

So, we convert into .

Let’s suppose that,

Let, in , angle and right angle at B.

[we will ignore the -ve sign because BC is a length and it can’t be -ve]

Inverse Trigonometric functions exercise 3.8 question 1 (viii)

As we know that the range of

In place of here is

So, we convert into .

Let’s suppose that,

Let, in , angle and right angle at B.

[we will ignore the -ve sign because BC is a length and it can’t be -ve]

Inverse Trigonometric functions exercise 3.8 question 1 (ix)

As we know that the range of

We have

In place of here is

So, we convert into .

Let’s suppose that,

Let, in , angle and right angle at B.

[we will ignore the -ve sign because AB is a length and it can’t be -ve]

Inverse Trigonometric functions exercise 3.8 question 2 (i)

We convert in the form of so we know the formula of it.

We have to prove that

LHS=

Let suppose that and ,

But we have not . We have only . So, we convert into .

Let, in , angle and right angle at B.

[ we take LCM of 3,4]

Hence proved.

Inverse trigonometric functions exercise 3.8 question 2 (ii)

We convert in the form of and we know the formula of it.

We have to prove

LHS =

Let’s suppose that and .

Here, we only have and .So we have to find and .

Let, in , angle and right angle at B.

Let, in , angle and right angle at B.

[we will ignore the -ve sign because PR is a length and it can’t be -ve]

Inverse trigonometric functions exercise 3.8 question 2 (iii)

We convert expression in the form of so we know the formula of it.

We have to prove

LHS =

…(i)

But we have not and . So, we find out first then we put in equation (i)

Let, in , angle and right angle at B.

[we will ignore the -ve sign because AB is a length and it can’t be -ve]

Let, in , angle and right angle at B.

Now, [ put, and ]

[ we take LCM of 3 and 12]

Hence proved.

Inverse trigonometric functions exercise 3.8 question 2 (iv)

We convert in the form of and we know the formula of it.

We have to prove

LHS =

Let’s suppose that

and

…(i)

Here we don’t have . So, we find out it and then put all the values in equation (i)

Let, in , angle and right angle at B.

Let, in , angle and right angle at B.

[we will ignore the -ve sign because PQ is a length and it can’t be -ve]

Now,

Hence proved.

Inverse trigromtery functions exercise 3.8 question 3 (i)

We know the value of

We have to solve to find the value of x.

Here, we have given in place of . So, we convert into

Let’s suppose that,

We use the identity,

[But we can eliminate negative value of , because in RHS of the question only positive value is given.]

…(i)

Now,

.....(ii)

From equation (i) and (ii) we get,

Squaring on both sides,

Therefore, the answer is .

Inverse trigromtery functions exercise 3.8 question 3 (ii)

**Hint:**

As we know that the value of**Given:**

We have to solve to find the value of x.**Solution:**

So, we convert into

[]

…(i) [ as, is even function, so ]

Let’s suppose,

…(ii)

…(iii)

Given,

… [ from eqn(i) ]

…[ using (ii) ]

[ ]

Therefore, the answer is

The 3rd chapter in mathematics, Inverse Trigonometry for Class 12, is not easy for many students to understand. Especially in the eighth exercise, which is Exercise 3.8, the students must solve the functions as learned from the previous exercises. The concepts given here are regarding solving sine, cosine, tangent, cotangent, secant, and cosecant functions. There are about fifteen questions, including the subparts in this exercise 3.8. If they find it hard, the RD Sharma Class 12 Chapter 3 Exercise 3.8 solutions will lend them a helping hand.

The solutions in this book are created by experts and verified according to the updated syllabus. You can use this guide to complete your homework and make your assignments too. As it follows the NCERT pattern, it is beneficial for the students who pursue their education at CBSE institutions. If you see yourself scoring low marks in the chapter Inverse Trigonometry, use the solutions given in the Class 12 RD Sharma Chapter 3 Exercise 3.8 Solution and understand the concepts. The different ways to solve a single question are given in the material. Soon, you will find yourself crossing your benchmark score.

Looking at the benefits, you attain from this RD Sharma book, you might presume that it might cost a lot. But that is not the fact. The RD Sharma Class 12 Solutions Inverse Trigonometry Ex 3.8 can be downloaded for free from top websites like Career360. All you need is to visit the Careers360 website and search for the solutions of Class 12 Mathematics Chapter 3 for any exercises that you need. You will be able to access the best set of RD Sharma solutions free of cost.

Due to the wide recognition of the solutions provided in the RD Sharma book, there are many chances that you will be asked questions from this book at your public examinations. Many students have benefitted by using the RD Sharma Class 12 Solutions Chapter 3 ex 3.8 to understand the concepts in Inverse Trigonometry deeply. Download your set of RD Sharma solutions book now from the Career360 website.

**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. Which guide is the best for class 12 students to prepare well for their public mathematics exam?

The RD Sharma books are a set of solutions provided by experts in the educational sector. Hence, the students can use it to prepare for their exams.

2. What to do if I have any doubts about the Inverse Trigonometry chapter?

If you are a CBSE board class 12 student, and the chapter Inverse Trigonometry seems to be a bit hard for you, get help from the RD Sharma Class 12th exercise 3.8 solutions book to understand the concepts better.

3. Can everyone access the RD Sharma Class 12th exercise 3.8 solution?

Anyone who needs a better solution material can access the RD Sharma solution at the Career360 website for free of cost

4. Is it enough if I work out the sums for chapter 3 given in the RD Sharma book?

It is more than enough to work out the sums in the RD Sharma Class 12 Chapter 3 Exercise 3.8 solutions for a deeper understanding of its mathematical concepts.

5. Are the solutions provided in RD Sharma Class 12th exercise 3.8 easy to understand?

With the solutions of educational experts, RD Sharma Class 12th exercise 3.8 solution consists of many methods of solving a mathematical function. Hence, both the toppers and the average students can find it easier to choose the way they understand and go ahead with it.

Mar 22, 2023

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