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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:Therangeoftheprincipalvalueofsec1is[0,π]{π2}
Given: sec1(2)
Solution:Let,y=sec1(2)
secy=2.
Weknowthatsecπ4=2
sec(π4)=2
sec(ππ4)
sec(3π4)
Thus,therangeoftheprincipalvalueofsec1is[0,π]{π2}
sec(3π4)=2Hence,theprincipalvalueofsec1(2)is3π4.


Inverse Trigonometric Function Exercise 3.4 Question 1 (ii)

Answer: π3
Hint: The range of principal value of sec1is[0,π]{π2}
Given:sec1(2)
Solution:Lety=sec1(2)
secy=2
As we knowsec(π3)=2
Therefore the range of principal value of sec1is[0,π]{π2}andsec(π3)=2
Thus the principal value of sec1(2)isπ3

Inverse Trigonometric Function Exercise 3.4 Question 1 (iii)

Answer:π4
Hint: The range of principal value of sec1is[0,π]{π2}
Given: sec1(2sin(3π4))
Solution: We know that sin(3π4)=12
2sin(3π4)2×12
2sin(3π4)2[2×2=2]
By substituting these values insec1(2sin(3π4)),weget
sec1(2)
Lety=2
sec(π4)=2
Therefore the range of principal value of sec1is[0,π]{π2}andsec(π4)=2
Thus,the principal value of sec1(2sin(3π4))is.π4

Inverse Trigonometric Function Exercise 3.4 Question 1 (iv)

Answer:2π3
Hint: The range of the principal value of sec1is[0,π]{π2}
Given: sec1(2tan(3π4))
Solution: We know that tan(3π4)=1
2tan(3π4)2×1
2tan(3π4)2
By substituting these values insec1(2tan(3π4)),we get
sec1(2)
Let ,y=sec1(2)
sec(y)=2
sec(π3)=2
sec(ππ3)
sec(2π3)
The range of the principal value of sec1is[0,π]{π2}andsec(2π3)=2
The principal value of sec1(2tan(3π4))is2π3

Inverse Trigonometric Function Exercise 3.4 Question 2 (i)

Answer:π3
Hint: The range of the principal value of sec1is[0,π]{π2}
The range of the principal value of tan1is[π2,π2]
Given:tan1(3)sec1(2)
Solution:First we solve tan1(3)
Let,y=tan1(3)
tany=(3)
tany=tan(π3)
Since[tan(π3)=3]
y=π3
Since range of tan1is[π2,π2]

Hence, Principal value is π3

Now, Solving sec1(2)

Let,y=sec1(2)

We know that

sec1(x)=πsec1(x)
y=πsec1(2) {secπ3=2π3=sec12}

y=ππ3
y=2π3
Since range of the principal value of sec1is[0,π]{π2}
Hence, principal value is 2π3
Now, we have
tan1(3)=π3andsec1(2)=2π3
Solving
tan1(3)sec1(2)
π32π3
π2π3
π3

Inverse Trigonometric Function Exercise 3.4 Question 2 (ii)

Answer:2π3
Hint:The range of the principal value of sec1is[0,π]{π2}
The range of the principal value of sin1is[π2,π2]
Given: sin1(32)2sec1(2tanπ6)
Solution: First we solve sin1(32)
Let y=sin1(32)
siny=(32)
sin(π3)
As we know sin(θ)=sin(θ)
sin(π3)=sin(π3)
The range of the principal value of sin1is[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 sec1(2tanπ6) we get
sec1(23)
Now let sec1(23)=z
secz=23
sec(π6)=23
The range of the principal value of sec1is[0,π]{π2}andsec(π3)=23
Hence, Principal value of sec1(2tan(π6))isπ3
Now, We have
sec1(2tan(π6))=π3andsin1(32)=π3
Solving
sin1(32)2sec1(2tan(π6))
π3π3
2π3

Inverse Trigonometric Function Exercise 3.4 Question 3 (i)

Answer:(,0][23,)
Hint:Domain(y1)=Range(y)
Domainofsec1(x)=(,1]or[1,]
Given: sec1(3x1)
Solution:Weknowthattherangeofsecxis(,1)(1,)
3x1<1and3x11
3x<1+1and3x1+1
3x<0and3x2
x<0andx23
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 sec1(x)is(,1][1,)
Domain of tan1(x) is R
Given: sec1(x)tan1x
Solution:We know that the domain of sec1(x)is(,1][1,)(i)
Domain of tan1xisR(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,).

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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.

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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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