RD Sharma Solutions Class 12 Chapter 4 Exercise 4.5 2022

RD Sharma Solutions Class 12 Chapter 4 Exercise 4.5

RD Sharma Solutions Class 12 Chapter 4 Exercise 4.5: The questions in Chapter 4’s Exercise 4.5 are based on the inverse of the cosecant function. Students who are unable to answer exercise wise questions as outlined in the RD Sharma textbook can use solutions created by Kopykitab’s expert faculty. The RD Sharma Solutions are presented in the most simple manner possible in order to provide students a thorough understanding of the concepts. It primarily increases their problem-solving speed, as well as their time management, which is critical for board exam success.

Students can download RD Sharma Solutions Class 12 Maths Chapter 4 Inverse Trigonometric Functions Exercise 4.5 PDF to get their doubts addressed straight away.

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Download RD Sharma Solutions Class 12 Chapter 4 Exercise 4.5 Free PDF

RD Sharma Solutions Class 12 Chapter 4 Exercise 4.5

Access answers to Maths RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.5 Important Questions With Solution

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 a 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

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

We have provided complete details of RD Sharma Solutions for Class 12 Maths Chapter 4 Exercise 4.5. If you have any queries related to CBSE Class 12 Exam, feel free to ask us in the comment section below.

FAQs on RD Sharma Solutions Class 12 Maths Chapter 4 Exercise 4.5

How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 4 Exercise 4.5?

There are a total of 9 questions in RD Sharma Solutions for Class 12 Maths Chapter 4 Exercise 4.5.

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Yes, You can get RD Sharma Solutions Class 12 Maths Chapter 4 Exercise 4.5 for free.

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RD Sharma Solutions For Class 12 Maths Exercise 4.5 Chapter 4 (Updated for 2021-22)