RD Sharma Solutions Class 12 Maths Exercise 4.8 Chapter 4 Inverse Trigonometric Functions are provided here. Exercise 4.8 of Chapter 4 deals with the properties of inverse functions. By using these properties, we have to solve the given problems. These concepts are explained in the best possible way by a team of experts at BYJU’S to help students prepare effectively for the board exam. The solutions are prepared in a step-wise 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.
RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.8
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)
(iii) Given Sin (tan-1 24/7)
(iv) Given Sin (sec-1 17/8)
(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)
(vii) Given Tan (cos-1 8/17)
(viii) Given cot (cos-1 3/5)
(ix) Given Cos (tan-1 24/7)
.
Comments