# RD Sharma Solutions For Class 12 Maths Exercise 4.4 Chapter 4 Inverse Trigonometric Functions

RD Sharma Solutions Class 12 Maths Exercise 4.4 Chapter 4 Inverse Trigonometric Functions is provided here. This exercise explains inverse of secant function and range lies in secant function. These concepts are explained in simple language by a set of expert faculty at BYJUâ€™S. The problems are solved in an interactive manner to make it interesting for the students while solving them.

The primary aim of preparing RD Sharma Solutions is to help students with their exam preparation. To score well in the exam, students can refer to RD Sharma Solutions Class 12 Maths Chapter 4 Inverse Trigonometric Functions Exercise 4.4 PDF, which are provided below.

## Download the PDF of RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.4

### Exercise 4.4 Page No: 4.18

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

(i) sec-1 (-âˆš2)

(ii) sec-1 (2)

(iii) sec-1 (2 sin (3Ï€/4))

(iv) sec-1 (2 tan (3Ï€/4))

Solution:

(i) Given sec-1 (-âˆš2)

Now let y = sec-1 (-âˆš2)

Sec y = -âˆš2

We know that sec Ï€/4 = âˆš2

Therefore, -sec (Ï€/4) = -âˆš2

= sec (Ï€ – Ï€/4)

= sec (3Ï€/4)

Thus the range of principal value of sec-1 is [0, Ï€] â€“ {Ï€/2}

And sec (3Ï€/4) = – âˆš2

Hence the principal value of sec-1 (-âˆš2) is 3Ï€/4

(ii) Given sec-1 (2)

Let y = sec-1 (2)

Sec y = 2

= Sec Ï€/3

Therefore the range of principal value of sec-1 is [0, Ï€] â€“ {Ï€/2} and sec Ï€/3 = 2

Thus the principal value of sec-1 (2) is Ï€/3

(iii) Given sec-1 (2 sin (3Ï€/4))

But we know that sin (3Ï€/4) = 1/âˆš2

Therefore 2 sin (3Ï€/4) = 2 Ã— 1/âˆš2

2 sin (3Ï€/4) = âˆš2

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

Sec-1 (âˆš2)

Let Sec-1 (âˆš2) = y

Sec y = âˆš2

Sec (Ï€/4) = âˆš2

Therefore range of principal value of sec-1 is [0, Ï€] â€“ {Ï€/2} and sec (Ï€/4) = âˆš2

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

(iv) Given sec-1 (2 tan (3Ï€/4))

But we know that tan (3Ï€/4) = -1

Therefore, 2 tan (3Ï€/4) = 2 Ã— -1

2 tan (3Ï€/4) = -2

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

Sec-1 (-2)

Now let y = Sec-1 (-2)

Sec y = – 2

– sec (Ï€/3) = -2

= sec (Ï€ â€“ Ï€/3)

= sec (2Ï€/3)

Therefore the range of principal value of sec-1 is [0, Ï€] â€“ {Ï€/2} and sec (2Ï€/3) = -2

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

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

Exercise 4.1 Solutions

Exercise 4.2 Solutions

Exercise 4.3 Solutions

Exercise 4.5 Solutions

Exercise 4.6 Solutions

Exercise 4.7 Solutions

Exercise 4.8 Solutions

Exercise 4.9 Solutions

Exercise 4.10 Solutions

Exercise 4.11 Solutions

Exercise 4.12 Solutions

Exercise 4.13 Solutions

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