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RD Sharma Class 12 Exercise 3.4 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 3.4 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 12:02 PM IST

RD Sharma Class 12 Chapter 3 Exercise 3.4 is the one-stop solution for every student facing problems in learning mathematics, especially inverse trigonometry. Students will be able to solve the exercise 3.4 easily. The concepts and solutions of this chapter are formatted in a way that students will find it easy to solve the questions in multiple ways. RD Sharma class 12th exercise 3.4 solutions are there to help a student to score and compete well in any exam.

RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Functions Excercise: 3.4

Inverse Trigonometric Function Exercise 3.4 Question 1 (i)

Answer:3π4
Hint:Therangeoftheprincipalvalueofsec−1⁡is[0,π]−{π2}
Given: sec−1⁡(−2)
Solution:Let,y=sec−1⁡(−2)
sec⁡y=−2.
Weknowthatsec⁡π4=2
⇒−sec⁡(π4)=−2
⇒sec⁡(π−π4)
⇒sec⁡(3π4)
Thus,therangeoftheprincipalvalueofsec−1⁡is[0,π]−{π2}
⇒sec⁡(3π4)=−2Hence,theprincipalvalueofsec−1⁡(2)is3π4.


Inverse Trigonometric Function Exercise 3.4 Question 1 (ii)

Answer: π3
Hint: The range of principal value of sec−1⁡is[0,π]−{π2}
Given:sec−1⁡(2)
Solution:Lety=sec−1⁡(2)
sec⁡y=2
As we knowsec⁡(π3)=2
Therefore the range of principal value of sec−1is[0,π]−{π2}andsec⁡(π3)=2
Thus the principal value of sec−1⁡(2)isπ3

Inverse Trigonometric Function Exercise 3.4 Question 1 (iii)

Answer:π4
Hint: The range of principal value of sec−1⁡is[0,π]−{π2}
Given: sec−1⁡(2sin⁡(3π4))
Solution: We know that sin⁡(3π4)=12
∴2sin⁡(3π4)⇒2×12
∴2sin⁡(3π4)⇒2[2×2=2]
By substituting these values insec−1⁡(2sin⁡(3π4)),weget
sec−1⁡(2)
Lety=2
sec⁡(π4)=2
Therefore the range of principal value of sec−1⁡is[0,π]−{π2}andsec⁡(π4)=2
Thus,the principal value of sec−1⁡(2sin⁡(3π4))is.π4

Inverse Trigonometric Function Exercise 3.4 Question 1 (iv)

Answer:2π3
Hint: The range of the principal value of sec−1⁡is[0,π]−{π2}
Given: sec−1⁡(2tan⁡(3π4))
Solution: We know that tan⁡(3π4)=−1
∴2tan⁡(3π4)⇒2×−1
2tan⁡(3π4)⇒−2
By substituting these values insec−1⁡(2tan⁡(3π4)),we get
sec−1⁡(−2)
Let ,y=sec−1⁡(−2)
sec⁡(y)=−2
−sec⁡(π3)=−2
⇒sec⁡(π−π3)
⇒sec⁡(2π3)
∴The range of the principal value of sec−1⁡is[0,π]−{π2}andsec⁡(2π3)=−2
∴The principal value of sec−1⁡(2tan⁡(3π4))is2π3

Inverse Trigonometric Function Exercise 3.4 Question 2 (i)

Answer:−π3
Hint: The range of the principal value of sec−1⁡is[0,π]−{π2}
The range of the principal value of tan−1⁡is[−π2,π2]
Given:tan−1⁡(3)−sec−1⁡(−2)
Solution:First we solve tan−1⁡(3)
Let,y=tan−1⁡(3)
tan⁡y=(3)
tan⁡y=tan⁡(π3)
Since[tan⁡(π3)=3]
∴y=π3
Since range of tan−1⁡is[−π2,π2]

Hence, Principal value is π3

Now, Solving sec−1⁡(−2)

Let,y=sec−1⁡(−2)

We know that

sec−1⁡(−x)=π−sec−1⁡(x)
y=π−sec−1⁡(2) {∵sec⁡π3=2π3=sec−1⁡2}

y=π−π3
y=2π3
Since range of the principal value of sec−1⁡is[0,π]−{π2}
Hence, principal value is 2π3
Now, we have
tan−1⁡(3)=π3andsec−1⁡(−2)=2π3
Solving
tan−1⁡(3)−sec−1⁡(−2)
⇒π3−2π3
⇒π−2π3
⇒−π3

Inverse Trigonometric Function Exercise 3.4 Question 2 (ii)

Answer:−2π3
Hint:The range of the principal value of sec−1⁡is[0,π]−{π2}
The range of the principal value of sin−1⁡is[−π2,π2]
Given: sin−1⁡(32)−2sec−1⁡(2tan⁡π6)
Solution: First we solve sin−1⁡(32)
Let y=sin−1⁡(32)
⇒sin⁡y=(32)
⇒−sin⁡(π3)
⇒As we know sin⁡(−θ)=−sin⁡(θ)
−sin⁡(π3)=sin⁡(−π3)
The range of the principal value of sin−1⁡is[−π2,π2]andsin⁡(−π3)=−32
Hence, Principal value is −π3
Now, we solve 2tan⁡(π6)
Let us assume 2tan⁡(π6)=θ
We know that tan⁡(π6)=13
2tan⁡(π6)=2×13
2tan⁡(π6)=23
Now by substituting these values in sec−1⁡(2tan⁡π6) we get
sec−1⁡(23)
Now let sec−1⁡(23)=z
sec⁡z=23
sec⁡(π6)=23
The range of the principal value of sec−1⁡is[0,π]−{π2}andsec⁡(π3)=23
Hence, Principal value of sec−1⁡(2tan⁡(π6))isπ3
Now, We have
sec−1⁡(2tan⁡(π6))=π3andsin−1⁡(32)=−π3
Solving
sin−1⁡(32)−2sec−1⁡(2tan⁡(π6))
⇒−π3−π3
⇒−2π3

Inverse Trigonometric Function Exercise 3.4 Question 3 (i)

Answer:(−∞,0]∪[23,∞)
Hint:Domain(y−1)=Range(y)
Domainofsec−1⁡(x)=(−∞,−1]or[1,∞]
Given: sec−1⁡(3x−1)
Solution:Weknowthattherangeofsec⁡xis(−∞,−1)∪(1,∞)
3x−1<−1and3x−1≥1
3x<−1+1and3x≥1+1
3x<0and3x≥2
x<0andx≥23
xϵ(−∞,0]andxϵ[23,∞]
xϵ(−∞,0]∪[23,∞]
Domainis(−∞,0]∪[23,∞]


Inverse Trigonometric Function Exercise 3.4 Question 3 (ii)

Answer:(−∞,−1]∪[1,∞)
Hint: Domain of sec−1⁡(x)is(−∞,−1]∪[1,∞)
Domain of tan−1⁡(x) is R
Given: sec−1⁡(x)−tan−1⁡x
Solution:We know that the domain of sec−1⁡(x)is(−∞,1]∪[1,∞)⋅⋅⋅⋅⋅(i)
Domain of tan−1⁡xisR⋅⋅⋅⋅⋅⋅(ii)
Union of (i) and (ii) will be domain of the given function
(−∞,−1]∪[1,∞)∪R
⇒(−∞,−1]∪[1,∞)
The domain of the given function is (−∞,−1]∪[1,∞).

Mathematics is considered a tough subject by many students, and these solutions help every student score well exceptionally and learn a lot. In addition, a wide variety of questions are given in the book to ensure that no student is behind in studies and performs well academically.

The solutions cover all the topics and also highlight the important topics from the exam point of view. RD Sharma Class 12 Chapter 3 Exercise 3.4 Solutions are formatted and clearly so that any student can learn it quickly. As a result, students frequently find a few concepts difficult to grasp.

Expert-created practice questions could help a student assess self-performance. Furthermore, the questions are structured as per the exam pattern, making it much easier for a student to understand the exam pattern much more effectively. The material in these solutions will be helpful for every exam as it is formulated by a team of experts. There are practice questions and examples which are written by a team of experts to make sure that a student learns every concept perfectly. RD Sharma class 12th exercise 3.4 has fulfilled all the requirements of a topic wonderfully to help every student.

Students can go through these solutions to clear the concepts and score brilliantly in exams. Every student should practice these solutions to get the benefits.The students must buy class 12 RD Sharma chapter 3 exercise 3.4 solution to have full access to learning and development.

RD Sharma Solutions Class 12 Exercise 3.4 includes level one having three questions along with their answers to practice and learn. Students will learn about Principle values and domains in this exercise and will solve these questions to better understand the concepts. The RD Sharma class 12 solutions chapter 3 ex 3.4 has everything a student requires to gain knowledge about the subject.

RD Sharma class 12 chapter 3 solutions are free of cost on Career360. Many students have taken advantage of these solutions created by Experts, and now it is your turn to take advantage of the exciting offer. If you're not sure about it, you can go through the website and find more expert-created answers.

Chapter-wise RD Sharma Class 12 Solutions

Frequently Asked Questions (FAQs)

1. What  are the six trigonometric functions?

Sine, Cosine, Tangent, Secant, Cosecant, and Cotangent are the six trigonometric functions

2. What are the three most fundamental trigonometric functions?

Sine, Cosine, and Tangent are the three fundamental trigonometric functions.

3. What is the value of sin θ, cos θ and tan θ, if θ=30 degrees?

If θ = 30 degrees, then,

Sin θ = sin 30 = ½

Cos θ = cos 30 = √3/2

Tan θ = tan 30 = 1/√3

4. Are the solutions provided in the RD Sharma class 12 books prepared from the exam point of view?

Yes, the solutions provided in the RD Sharma class 12 books are prepared from the exam point of view as these solutions are prepared by experts and are constructed from the exam point of view.

5. Is the RD Sharma Chapter 3 Solutions book enough for the CBSE board students to prepare for their public exams?

The RD Sharma Solutions books are more than enough for the class 12 students to prepare for their board exams. In addition, these books provide a deeper insight into the concepts to the students hence making it easier for them to learn.

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