RD Sharma Solutions For Class 12 Maths Exercise 4.8 Chapter 4 Inverse Trigonometric Functions

RD Sharma Solutions Class 12 Maths Exercise 4.8 Chapter 4 Inverse Trigonometric Functions is provided here. Exercise 4.8 of Chapter 4 deals with the properties of inverse functions. By using these properties we have to solve given problems. These concepts are explained in the best possible way by a set of experts at BYJU’S to help students prepare effectively for the board exam. The solutions are prepared in a stepwise manner as each step carries marks in the latest CBSE exam pattern.

Students can refer to the PDF of RD Sharma Solutions while solving exercise wise problems to get their doubts cleared immediately. RD Sharma Solutions Class 12 Maths Chapter 4 Inverse Trigonometric Functions Exercise 4.8 PDF are available here.

Download the PDF of RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.8

 

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Access answers to Maths RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.8

Exercise 4.8 Page No: 4.54

1. Evaluate each of the following:

(i) sin (sin-1 7/25)

(ii) Sin (cos-1 5/13)

(iii) Sin (tan-1 24/7)

(iv) Sin (sec-1 17/8)

(v) Cosec (cos-1 8/17)

(vi) Sec (sin-1 12/13)

(vii) Tan (cos-1 8/17)

(viii) cot (cos-1 3/5)

(ix) Cos (tan-1 24/7)

Solution:

(i) Given sin (sin-1 7/25)

Now let y = sin-1 7/25

Sin y = 7/25 where y ∈ [0, π/2]

Substituting these values in sin (sin-1 7/25) we get

Sin (sin-1 7/25) = 7/25

(ii) Given Sin (cos-1 5/13)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 18

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 19

(iii) Given Sin (tan-1 24/7)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 20

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 21

(iv) Given Sin (sec-1 17/8)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 22

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 23

(v) Given Cosec (cos-1 8/17)

Let cos-1(8/17) = y

cos y = 8/17 where y ∈ [0, π/2]

Now, we have to find

Cosec (cos-1 8/17) = cosec y

We know that,

sin2 θ + cos2 θ = 1

sin2 θ = √ (1 – cos2 θ)

So,

sin y = √ (1 – cos2 y)

= √ (1 – (8/17)2)

= √ (1 – 64/289)

= √ (289 – 64/289)

= √ (225/289)

= 15/17

Hence,

Cosec y = 1/sin y = 1/ (15/17) = 17/15

Therefore,

Cosec (cos-1 8/17) = 17/15

(vi) Given Sec (sin-1 12/13)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 26

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 27

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 28

(vii) Given Tan (cos-1 8/17)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 29

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 30

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 31

(viii) Given cot (cos-1 3/5)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 32

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 33

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 34

(ix) Given Cos (tan-1 24/7)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 35.

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 36


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 4.4 Solutions

Exercise 4.5 Solutions

Exercise 4.6 Solutions

Exercise 4.7 Solutions

Exercise 4.9 Solutions

Exercise 4.10 Solutions

Exercise 4.11 Solutions

Exercise 4.12 Solutions

Exercise 4.13 Solutions

Exercise 4.14 Solutions

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