RD Sharma Class 12 Solutions Inverse Trigonometric Functions PDF Download

RD Sharma Solutions for Class 12 Maths Chapter 4 – Inverse Trigonometric Functions are presented in a clear and comprehensive format to help students score well in CBSE board exams. As Class 12 plays a pivotal role in shaping academic and career goals, these solutions are designed with student-friendly language to match diverse learning needs.

Our expert faculty has developed exercise-wise solutions following the RD Sharma textbook structure, ensuring each concept is explained step by step for easy understanding. Students can download the RD Sharma Class 12 Solutions PDF for Chapter 4 to clarify doubts, solve problems with confidence, and enhance their exam preparation.

These RD Sharma solutions align with the latest CBSE syllabus, updated exam pattern, and current marking scheme for 2025-26. Regular practice of this chapter enables students to solve complex problems efficiently and boosts conceptual clarity. Let us check at some of the important concepts that are discussed in this chapter.

• Definition and meaning of inverse trigonometric functions
• Inverse of sine, cosine, and tangent functions
• Inverse of secant, cosecant, and cotangent functions
• Properties of inverse trigonometric functions

With 14 well-structured exercises, this chapter builds a strong foundation in understanding inverse trigonometric concepts essential for competitive exams as well.

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

RD Sharma Class 12 Maths Solutions Chapter 4 – Inverse Trigonometric Functions cover all the questions from the textbook, crafted by expert Mathematics teachers at Infinity Learn. Download our free PDF of Chapter 4 – Inverse Trigonometric Functions RD Sharma Solutions for Class 12 to boost your performance in board exams and competitive exams.

Access answers to RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions

Q. Find the domain of f(x) = cos^-1 2x + sin^-1 x.

Solution:

Given that f(x) = cos^-1 2x + sin^-1 x.

Now we have to find the domain of f(x),

We know that domain of cos^-1 x lies in the interval [-1, 1]

Also know that domain of sin^-1 x lies in the interval [-1, 1]

Therefore, the domain of cos^-1 (2x) lies in the interval [-1, 1]

Hence we can write as,

-1 ≤ 2x ≤ 1

– ½ ≤ x ≤ ½

Hence, domain of cos^-1(2x) + sin^-1 x lies in the interval [- ½, ½]

Q. Find the domain of definition of f(x) = cos ^-1 (x² – 4)

Solution: Given f(x) = cos ^-1 (x² – 4)

We know that domain of cos^-1 (x² – 4) lies in the interval [-1, 1]

Therefore, we can write as

-1 ≤ x² – 4 ≤ 1

4 – 1 ≤ x² ≤ 1 + 4

3 ≤ x² ≤ 5

±√ 3 ≤ x ≤ ±√5

– √5 ≤ x ≤ – √3 and √3 ≤ x ≤ √5

Therefore domain of cos^-1 (x² – 4) is [- √5, – √3] ∪ [√3, √5]

Q. 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

Q. 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 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

Q. Evaluate each of the following:

(i) sin^-1(sin π/6)

(ii) sin^-1(sin 7π/6)

(iii) sin^-1(sin 5π/6)

(iv) sin^-1(sin 13π/7)

(v) sin^-1(sin 17π/8)

Solution:

(i) Given sin^-1(sin π/6)

We know that the value of sin π/6 is ½

By substituting this value in sin^-1(sin π/6)

We get, sin^-1 (1/2)

Now let y = sin^-1 (1/2)

Sin (π/6) = ½

The range of principal value of sin^-1(-π/2, π/2) and sin (π/6) = ½

Therefore sin^-1(sin π/6) = π/6

(ii) Given sin^-1(sin 7π/6)

But we know that sin 7π/6 = – ½

By substituting this in sin^-1(sin 7π/6) we get,

Sin^-1 (-1/2)

Now let y = sin^-1 (-1/2)

– Sin y = ½

– Sin (π/6) = ½

– Sin (π/6) = sin (- π/6)

The range of principal value of sin^-1(-π/2, π/2) and sin (- π/6) = – ½

Therefore sin^-1(sin 7π/6) = – π/6

(iii) Given sin^-1(sin 5π/6)

We know that the value of sin 5π/6 is ½

By substituting this value in sin^-1(sin 5π/6)

We get, sin^-1 (1/2)

Now let y = sin^-1 (1/2)

Sin (π/6) = ½

The range of principal value of sin^-1(-π/2, π/2) and sin (π/6) = ½

Therefore sin^-1(sin 5π/6) = π/6

(iv) Given sin^-1(sin 13π/7)

Given question can be written as sin (2π – π/7)

Sin (2π – π/7) can be written as sin (-π/7) [since sin (2π – θ) = sin (-θ)]

By substituting these values in sin^-1(sin 13π/7) we get sin^-1(sin – π/7)

As sin^-1(sin x) = x with x ∈ [-π/2, π/2]

Therefore sin^-1(sin 13π/7) = – π/7

(v) Given sin^-1(sin 17π/8)

Given question can be written as sin (2π + π/8)

Sin (2π + π/8) can be written as sin (π/8)

By substituting these values in sin^-1(sin 17π/8) we get sin^-1(sin π/8)

As sin^-1(sin x) = x with x ∈ [-π/2, π/2]

Therefore sin^-1(sin 17π/8) = π/8

Q. Evaluate each of the following:

(i) cos^-1{cos (-π/4)}

(ii) cos^-1(cos 5π/4)

(iii) cos^-1(cos 4π/3)

(iv) cos^-1(cos 13π/6)

Solution:

(i) Given cos^-1{cos (-π/4)}

We know that cos (-π/4) = cos (π/4) [since cos (-θ) = cos θ

Also know that cos (π/4) = 1/√2

By substituting these values in cos^-1{cos (-π/4)} we get,

Cos^-1(1/√2)

Now let y = cos^-1(1/√2)

Therefore cos y = 1/√2

Hence range of principal value of cos^-1 is [0, π] and cos (π/4) = 1/√2

Therefore cos^-1{cos (-π/4)} = π/4

(ii) Given cos^-1(cos 5π/4)

But we know that cos (5π/4) = -1/√2

By substituting these values in cos^-1{cos (5π/4)} we get,

Cos^-1(-1/√2)

Now let y = cos^-1(-1/√2)

Therefore cos y = – 1/√2

– Cos (π/4) = 1/√2

Cos (π – π/4) = – 1/√2

Cos (3 π/4) = – 1/√2

Hence range of principal value of cos^-1 is [0, π] and cos (3π/4) = -1/√2

Therefore cos^-1{cos (5π/4)} = 3π/4

(iii) Given cos^-1(cos 4π/3)

But we know that cos (4π/3) = -1/2

By substituting these values in cos^-1{cos (4π/3)} we get,

Cos^-1(-1/2)

Now let y = cos^-1(-1/2)

Therefore cos y = – 1/2

– Cos (π/3) = 1/2

Cos (π – π/3) = – 1/2

Cos (2π/3) = – 1/2

Hence range of principal value of cos^-1 is [0, π] and cos (2π/3) = -1/2

Therefore cos^-1{cos (4π/3)} = 2π/3

(iv) Given cos^-1(cos 13π/6)

But we know that cos (13π/6) = √3/2

By substituting these values in cos^-1{cos (13π/6)} we get,

Cos^-1(√3/2)

Now let y = cos^-1(√3/2)

Therefore cos y = √3/2

Cos (π/6) = √3/2

Hence range of principal value of cos^-1 is [0, π] and cos (π/6) = √3/2

Therefore cos^-1{cos (13π/6)} = π/6