RD Sharma Solutions For Class 12 Maths Exercise 4.3 Chapter 4 (Updated for 2021-22)

RD Sharma Solutions Class 12 Chapter 4 Exercise 4.3: Students will be introduced to the problems in order to assist them analyse their understanding of the chapter. Under RD Sharma Solutions exercise 4.3, the various steps involved in solving questions with inverse trigonometric functions are briefly explained. Students can use the solutions primarily as a reference tool to help them prepare for their board exams faster. Students can obtain RD Sharma Solutions Class 12 Maths Chapter 4 Inverse Trigonometric Functions Exercise 4.3 from the links provided below to boost their self-analytic capacity. 

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RD Sharma Solutions Class 12 Chapter 4 Exercise 4.3

Access answers to Maths RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.3 Important Questions With Solution

1. Find the principal value of each of the following:

(i) tan^-1 (1/√3)

(ii) tan^-1 (-1/√3)

(iii) tan^-1 (cos (π/2))

(iv) tan^-1 (2 cos (2π/3))

Solution:

(i) Given tan^-1 (1/√3)

We know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

So, tan^-1 (1/√3) = an angle in (-π/2, π/2) whose tangent is (1/√3)

But we know that the value is equal to π/6

Therefore tan^-1 (1/√3) = π/6

Hence the principal value of tan^-1 (1/√3) = π/6

(ii) Given tan^-1 (-1/√3)

We know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

So, tan^-1 (-1/√3) = an angle in (-π/2, π/2) whose tangent is (1/√3)

But we know that the value is equal to -π/6

Therefore tan^-1 (-1/√3) = -π/6

Hence the principal value of tan^-1 (-1/√3) = – π/6

(iii) Given that tan^-1 (cos (π/2))

But we know that cos (π/2) = 0

We know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

Therefore tan^-1 (0) = 0

Hence the principal value of tan^-1 (cos (π/2) is 0.

(iv) Given that tan^-1 (2 cos (2π/3))

But we know that cos π/3 = 1/2

So, cos (2π/3) = -1/2

Therefore tan^-1 (2 cos (2π/3)) = tan^-1 (2 × – ½)

= tan^-1(-1)

= – π/4

Hence, the principal value of tan^-1 (2 cos (2π/3)) is – π/4

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

• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.1
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.2
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.4
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.5
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.6
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.7
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.8
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.9
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.10
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.11
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.12
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.13
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.14

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