RD Sharma Solutions Class 12 Maths Exercise 4.5 Chapter 4 Inverse Trigonometric Functions are provided here. Exercise 4.5 of Chapter 4 consists of problems based on the inverse of the cosecant function. Students who are unable to solve exercise-wise problems as per the RD Sharma textbook can make use of solutions designed by expert faculty at BYJU’S.
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RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.5
Access answers to Maths RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.5
Exercise 4.5 Page No: 4.21
1. Find the principal values of each of the following:
(i) cosec-1 (-√2)
(ii) cosec-1 (-2)
(iii) cosec-1 (2/√3)
(iv) cosec-1 (2 cos (2π/3))
Solution:
(i) Given cosec-1 (-√2)
Let y = cosec-1 (-√2)
Cosec y = -√2
– Cosec y = √2
– Cosec (π/4) = √2
– Cosec (π/4) = cosec (-π/4) [since –cosec θ = cosec (-θ)]
The range of principal value of cosec-1 [-π/2, π/2] – {0} and cosec (-π/4) = – √2
Cosec (-π/4) = – √2
Therefore, the principal value of cosec-1 (-√2) is – π/4
(ii) Given cosec-1 (-2)
Let y = cosec-1 (-2)
Cosec y = -2
– Cosec y = 2
– Cosec (π/6) = 2
– Cosec (π/6) = cosec (-π/6) [since –cosec θ = cosec (-θ)]
The range of principal value of cosec-1 [-π/2, π/2] – {0} and cosec (-π/6) = – 2
Cosec (-π/6) = – 2
Therefore, the principal value of cosec-1 (-2) is – π/6
(iii) Given cosec-1 (2/√3)
Let y = cosec-1 (2/√3)
Cosec y = (2/√3)
Cosec (π/3) = (2/√3)
Therefore, range of principal value of cosec-1 is [-π/2, π/2] – {0} and cosec (π/3) = (2/√3)
Thus, the principal value of cosec-1 (2/√3) is π/3
(iv) Given cosec-1 (2 cos (2π/3))
But we know that cos (2π/3) = – ½
Therefore, 2 cos (2π/3) = 2 × – ½
2 cos (2π/3) = -1
By substituting these values in cosec-1 (2 cos (2π/3)) we get,
Cosec-1 (-1)
Let y = cosec-1 (-1)
– Cosec y = 1
– Cosec (π/2) = cosec (-π/2) [since –cosec θ = cosec (-θ)]
The range of principal value of cosec-1 [-π/2, π/2] – {0} and cosec (-π/2) = – 1
Cosec (-π/2) = – 1
Therefore, the principal value of cosec-1 (2 cos (2π/3)) is – π/2
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