RD Sharma Solutions for Class 12 Maths Chapter 4 – Inverse Trigonometric Functions

The RD Sharma Solutions for Class 12 Maths Chapter 4 – Inverse Trigonometric Functions are available here in PDF format. Students who face challenges while solving the exercise problems in this chapter can easily download the solutions PDF for RD Sharma Solutions. These solutions are prepared by expert maths tutors, strictly following the latest CBSE guidelines and updated marking schemes. The solutions are presented step-by-step to help students understand the concepts more clearly.

The RD Sharma Solutions for Class 12 are designed in a simple and easy-to-understand manner, aligning perfectly with the CBSE Class 12 Maths syllabus. Students preparing for exams can use these PDFs as a helpful study tool to boost their preparation. To understand the concepts discussed in this chapter more clearly, students can download the RD Sharma Solutions for Class 12 Chapter 4 Inverse Trigonometric Functions PDF from the links below.

RD Sharma Solutions for Class 12 – Inverse Trigonometric Functions

Inverse trigonometric functions are used to find an angle when the value of its trigonometric ratio is given. These functions are important in higher-level math, such as calculus and algebra. The RD Sharma Solutions for Class 12 introduces students to the definitions, properties, and main values of inverse trigonometric functions. The chapter explains how to solve and simplify expressions involving these functions with easy-to-understand examples.

Download RD Sharma Class 12 Chapter Inverse Trigonometric Functions PDF Solutions

Students can easily download the RD Sharma Class 12 Maths Chapter Inverse Trigonometric Functions Solutions PDF. It covers all exercises, solved examples, and additional practice questions. These PDF solutions are perfect for quick revision and help students gain a clear understanding of inverse trigonometric functions, making them ready for both board exams and competitive tests.

RD Sharma Solutions for Inverse Trigonometric Functions: Solved Questions

These RD Sharma Solutions for Class 12 will help students practice and get better at solving problems related to inverse trigonometric functions. Use these solutions to improve your skills and prepare confidently for exams.

Question 1: Define inverse trigonometric functions

Definition: Inverse trigonometric functions are the inverse functions of the trigonometric functions. They are used to find the angle when the value of the trigonometric ratio is given.

Examples:

1. sin⁻¹(x) or arcsin(x):

• Domain: [-1, 1]
• Range (Principal value branch): [-π/2, π/2]
• Example: sin⁻¹(1/2) = π/6

2. cos⁻¹(x) or arccos(x):

• Domain: [-1, 1]
• Range (Principal value branch): [0, π]
• Example: cos⁻¹(1/2) = π/3

Question 2: Solve sin⁻¹(1/2) = x

Solution:

We need to find the angle x such that sin(x) = 1/2 within the principal value range.

We know that sin(π/6) = 1/2

Since π/6 lies in the range [-π/2, π/2], we have:

x = π/6

Range of values for x: x ∈ [-π/2, π/2]

Question 3: Simplify tan⁻¹(1) + tan⁻¹(2)

Solution:

Using the addition formula: tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1-ab)) when ab < 1

Here, a = 1, b = 2

ab = 1 × 2 = 2 > 1, so we use the modified formula:

tan⁻¹(a) + tan⁻¹(b) = π + tan⁻¹((a+b)/(1-ab)) when ab > 1 and a,b > 0

tan⁻¹(1) + tan⁻¹(2) = π + tan⁻¹((1+2)/(1-1×2))

= π + tan⁻¹(3/(-1))

= π + tan⁻¹(-3)

= π - tan⁻¹(3)

Answer: π - tan⁻¹(3)

Question 4: If cos⁻¹(x) = π/3, find x

Solution:

If cos⁻¹(x) = π/3, then by definition:

cos(π/3) = x

We know that cos(π/3) = 1/2

Therefore, x = 1/2

Question 5: Prove sin⁻¹x + cos⁻¹x = π/2 for x ∈ [-1,1]

Proof:

Let sin⁻¹x = α, where α ∈ [-π/2, π/2]

This means sin(α) = x

We know that sin(α) = cos(π/2 - α)

So, cos(π/2 - α) = x

Taking cos⁻¹ on both sides:

π/2 - α = cos⁻¹x

Therefore: cos⁻¹x = π/2 - α = π/2 - sin⁻¹x

Rearranging: sin⁻¹x + cos⁻¹x = π/2

Hence proved.

Question 6: Find the principal value of cot⁻¹(-√3)

Solution:

We need to find θ such that cot(θ) = -√3 and θ ∈ (0, π)

We know that cot(π/6) = √3

Since cot(θ) = -√3, we need an angle in the second quadrant where cotangent is negative.

cot(π - π/6) = cot(5π/6) = -cot(π/6) = -√3

Therefore, cot⁻¹(-√3) = 5π/6

Question 7: Express tan⁻¹(3/4) in terms of sine function

Solution:

Let tan⁻¹(3/4) = θ

This means tan(θ) = 3/4

In a right triangle where opposite = 3 and adjacent = 4:

Hypotenuse = √(3² + 4²) = √(9 + 16) = √25 = 5

Therefore: sin(θ) = opposite/hypotenuse = 3/5

So: θ = sin⁻¹(3/5)

Therefore, tan⁻¹(3/4) = sin⁻¹(3/5)

Question 8: Solve 2sin⁻¹x = π/2 where x ∈ [-1,1]

Solution:

2sin⁻¹x = π/2

sin⁻¹x = π/4

Taking sine on both sides:

x = sin(π/4)

x = 1/√2 = √2/2

Therefore, x = √2/2

Question 9: Simplify 2tan⁻¹(x) and state valid range

Solution:

Using the double angle formula for inverse tangent:

2tan⁻¹(x) = tan⁻¹(2x/(1-x²)) when |x| < 1

2tan⁻¹(x) = π + tan⁻¹(2x/(1-x²)) when x > 1

2tan⁻¹(x) = -π + tan⁻¹(2x/(1-x²)) when x < -1

Valid ranges:

• For |x| < 1: 2tan⁻¹(x) = tan⁻¹(2x/(1-x²))
• For x > 1: 2tan⁻¹(x) = π + tan⁻¹(2x/(1-x²))
• For x < -1: 2tan⁻¹(x) = -π + tan⁻¹(2x/(1-x²))

Question 10: If y = sin⁻¹(cos x) for x ∈ [0,π], find y in terms of x

Solution:

We know that cos x = sin(π/2 - x)

So: y = sin⁻¹(cos x) = sin⁻¹(sin(π/2 - x))

For x ∈ [0, π/2]: π/2 - x ∈ [0, π/2]

Since π/2 - x is in the range of sin⁻¹, we have:

y = π/2 - x

For x ∈ [π/2, π]: π/2 - x ∈ [-π/2, 0]

Since π/2 - x is in the range of sin⁻¹, we have:

y = π/2 - x

Therefore, for x ∈ [0, π]: y = π/2 - x