RD Sharma Solutions For Class 12 Maths Exercise 4.4 Chapter 4 (Updated for 2021-22)

RD Sharma Solutions Class 12 Chapter 4 Exercise 4.4: This exercise illustrates what the inverse of a secant function is and where the range of a secant function is. At Kopykitab, a team of expert faculty members explain these RD Sharma Solutions concepts in simple language. The questions are solved in an interactive manner to keep the students engaged while they work on them. Preparing solutions are mostly intended to assist students in their board exam preparation. Students can use the RD Sharma Solutions Class 12 Maths Chapter 4 Inverse Trigonometric Functions Exercise 4.4 PDF to help them do well on the exam.

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RD Sharma Solutions Class 12 Chapter 4 Exercise 4.4

Access answers to Maths RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.4 Important Questions With Solution

1. Find the principal value of each of the following:

(i) sec^-1 (-√2)

(ii) sec^-1 (2)

(iii) sec^-1 (2 sin (3π/4))

(iv) sec^-1 (2 tan (3π/4))

Solution:

(i) Given sec^-1 (-√2)

Now let y = sec^-1 (-√2)

Sec y = -√2

We know that sec π/4 = √2

Therefore, -sec (π/4) = -√2

= sec (π – π/4)

= sec (3π/4)

Thus the range of principal value of sec^-1 is [0, π] – {π/2}

And sec (3π/4) = – √2

Hence the principal value of sec^-1 (-√2) is 3π/4

(ii) Given sec^-1 (2)

Let y = sec^-1 (2)

Sec y = 2

= Sec π/3

Therefore the range of principal value of sec^-1 is [0, π] – {π/2} and sec π/3 = 2

Thus the principal value of sec^-1 (2) is π/3

(iii) Given sec^-1 (2 sin (3π/4))

But we know that sin (3π/4) = 1/√2

Therefore 2 sin (3π/4) = 2 × 1/√2

2 sin (3π/4) = √2

Therefore by substituting above values in sec^-1 (2 sin (3π/4)), we get

Sec^-1 (√2)

Let Sec^-1 (√2) = y

Sec y = √2

Sec (π/4) = √2

Therefore range of principal value of sec^-1 is [0, π] – {π/2} and sec (π/4) = √2

Thus the principal value of sec^-1 (2 sin (3π/4)) is π/4.

(iv) Given sec^-1 (2 tan (3π/4))

But we know that tan (3π/4) = -1

Therefore, 2 tan (3π/4) = 2 × -1

2 tan (3π/4) = -2

By substituting these values in sec^-1 (2 tan (3π/4)), we get

Sec^-1 (-2)

Now let y = Sec^-1 (-2)

Sec y = – 2

– sec (π/3) = -2

= sec (π – π/3)

= sec (2π/3)

Therefore the range of principal value of sec^-1 is [0, π] – {π/2} and sec (2π/3) = -2

Thus, the principal value of sec^-1 (2 tan (3π/4)) is (2π/3).

Access other exercises of RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions

• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.1
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.2
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.3
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.5
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.6
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.7
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.8
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.9
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.10
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.11
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.12
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.13
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.14

We have provided complete details of RD Sharma Solutions Class 12 Chapter 4 Exercise 4.4. If you have any queries related to CBSE Class 12 Exam, feel free to ask us in the comment section below.

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