RD Sharma Solutions Class 12 Maths Exercise 4.6 Chapter 4 Inverse Trigonometric Functions are provided here. Students can refer to the solutions PDF while solving exercise-wise problems. This helps them to understand the other possible ways of obtaining answers easily. The solutions are designed by subject-matter experts based on the grasping abilities of students. It improves logical thinking skills among students, which are important for the board exam.
This exercise is completely based on the inverse of the cotangent function. Students who aim to clear the exam with good marks can use RD Sharma Solutions Class 12 Maths Chapter 4 Inverse Trigonometric Functions Exercise 4.6 PDF as a reference guide to boost their exam preparation.
RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.6
Access answers to Maths RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.6
Exercise 4.6 Page No: 4.24
1. Find the principal values of each of the following:
(i) cot-1(-√3)
(ii) Cot-1(√3)
(iii) cot-1(-1/√3)
(iv) cot-1(tan 3π/4)
Solution:
(i) Given cot-1(-√3)
Let y = cot-1(-√3)
– Cot (π/6) = √3
= Cot (π – π/6)
= cot (5π/6)
The range of principal value of cot-1 is (0, π) and cot (5 π/6) = – √3
Thus, the principal value of cot-1 (- √3) is 5π/6
(ii) Given Cot-1(√3)
Let y = cot-1(√3)
Cot (π/6) = √3
The range of principal value of cot-1 is (0, π) and
Thus, the principal value of cot-1 (√3) is π/6
(iii) Given cot-1(-1/√3)
Let y = cot-1(-1/√3)
Cot y = (-1/√3)
– Cot (π/3) = 1/√3
= Cot (π – π/3)
= cot (2π/3)
The range of principal value of cot-1(0, π) and cot (2π/3) = – 1/√3
Therefore, the principal value of cot-1(-1/√3) is 2π/3
(iv) Given cot-1(tan 3π/4)
But we know that tan 3π/4 = -1
By substituting this value in cot-1(tan 3π/4) we get
Cot-1(-1)
Now, let y = cot-1(-1)
Cot y = (-1)
– Cot (π/4) = 1
= Cot (π – π/4)
= cot (3π/4)
The range of principal value of cot-1(0, π) and cot (3π/4) = – 1
Therefore, the principal value of cot-1(tan 3π/4) is 3π/4
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