RD Sharma Solutions Class 12 Maths Exercise 4.5 Chapter 4 Inverse Trigonometric Functions is provided here. Exercise 4.5 of Chapter 4 consists of problems based on inverse of cosecant function. Students who are unable to solve exercise wise problems as per RD Sharma textbook can make use of solutions designed by expert faculty at BYJU’S.
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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 (2Ï€/3))
Solution:
(i) Given cosec-1 (-√2)
Let y = cosec-1 (-√2)
Cosec y = -√2
– Cosec y = √2
– Cosec (Ï€/4) = √2
– Cosec (Ï€/4) = cosec (-Ï€/4) [since –cosec θ = cosec (-θ)]
The range of principal value of cosec-1 [-Ï€/2, Ï€/2] – {0} and cosec (-Ï€/4) = – √2
Cosec (-Ï€/4) = – √2
Therefore the principal value of cosec-1 (-√2) is – Ï€/4
(ii) Given cosec-1 (-2)
Let y = cosec-1 (-2)
Cosec y = -2
– Cosec y = 2
– Cosec (Ï€/6) = 2
– Cosec (Ï€/6) = cosec (-Ï€/6) [since –cosec θ = cosec (-θ)]
The range of principal value of cosec-1 [-Ï€/2, Ï€/2] – {0} and cosec (-Ï€/6) = – 2
Cosec (-Ï€/6) = – 2
Therefore the principal value of cosec-1 (-2) is – Ï€/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 (2Ï€/3))
But we know that cos (2Ï€/3) = – ½
Therefore 2 cos (2Ï€/3) = 2 × – ½
2 cos (2Ï€/3) = -1
By substituting these values in cosec-1 (2 cos (2Ï€/3)) we get,
Cosec-1 (-1)
Let y = cosec-1 (-1)
– Cosec y = 1
– Cosec (Ï€/2) = cosec (-Ï€/2) [since –cosec θ = cosec (-θ)]
The range of principal value of cosec-1 [-Ï€/2, Ï€/2] – {0} and cosec (-Ï€/2) = – 1
Cosec (-Ï€/2) = – 1
Therefore the principal value of cosec-1 (2 cos (2Ï€/3)) is – Ï€/2
