RD Sharma Solutions Class 12 Maths Exercise 4.4 Chapter 4 Inverse Trigonometric Functions are provided here. This exercise explains the inverse of the secant function, and the range lies in the secant function. These concepts are explained in simple language by a team of expert faculty at BYJU’S. The problems are solved in an interactive manner to make it interesting for the students while solving them.
The primary aim of preparing RD Sharma Solutions is to help students with their exam preparation. To score well in the exam, students can refer to RD Sharma Solutions Class 12 Maths Chapter 4 Inverse Trigonometric Functions Exercise 4.4 PDF, which are provided below.
RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.4
Access answers to Maths RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.4
Exercise 4.4 Page No: 4.18
1. Find the principal value of each of the following:
(i) sec-1 (-√2)
(ii) sec-1 (2)
(iii) sec-1 (2 sin (3π/4))
(iv) sec-1 (2 tan (3π/4))
Solution:
(i) Given sec-1 (-√2)
Now let y = sec-1 (-√2)
Sec y = -√2
We know that sec π/4 = √2
Therefore, -sec (π/4) = -√2
= sec (π – π/4)
= sec (3π/4)
Thus, the range of principal value of sec-1 is [0, π] – {π/2}
And sec (3π/4) = – √2
Hence, the principal value of sec-1 (-√2) is 3π/4
(ii) Given sec-1 (2)
Let y = sec-1 (2)
Sec y = 2
= Sec π/3
Therefore, the range of principal value of sec-1 is [0, π] – {π/2} and sec π/3 = 2
Thus, the principal value of sec-1 (2) is π/3
(iii) Given sec-1 (2 sin (3π/4))
But we know that sin (3π/4) = 1/√2
Therefore, 2 sin (3π/4) = 2 × 1/√2
2 sin (3π/4) = √2
Therefore, by substituting the above values in sec-1 (2 sin (3π/4)), we get
Sec-1 (√2)
Let Sec-1 (√2) = y
Sec y = √2
Sec (π/4) = √2
Therefore, range of principal value of sec-1 is [0, π] – {π/2} and sec (π/4) = √2
Thus, the principal value of sec-1 (2 sin (3π/4)) is π/4.
(iv) Given sec-1 (2 tan (3π/4))
But we know that tan (3π/4) = -1
Therefore, 2 tan (3π/4) = 2 × -1
2 tan (3π/4) = -2
By substituting these values in sec-1 (2 tan (3π/4)), we get
Sec-1 (-2)
Now let y = Sec-1 (-2)
Sec y = – 2
– sec (π/3) = -2
= sec (π – π/3)
= sec (2π/3)
Therefore, the range of principal value of sec-1 is [0, π] – {π/2} and sec (2π/3) = -2
Thus, the principal value of sec-1 (2 tan (3π/4)) is (2π/3).
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