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

RD Sharma Solutions Class 12 Maths Chapter 4 – Free PDF Download Updated for (2023-24)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions are provided in a comprehensive manner for students to secure good marks in the board exam. Class 12 is a crucial stage in a student’s life, as it helps them achieve their career goals. We mainly focus on providing answers, keeping in mind the grasping abilities of students. To make learning fun and interesting, our team of faculty has provided exercise-wise solutions for each chapter as per the RD Sharma Solutions. Students can download the solutions and use them to solve problems and get their doubts cleared instantly. 

The solutions are according to the latest CBSE syllabus, the exam pattern and mark weightage in a step-wise manner. RD Sharma Solutions for Class 12 Chapter 4 Inverse Trigonometric Functions PDF are given here. Regular practice of Chapter 4 Inverse Trigonometric Functions boosts confidence to solve complex problems in an efficient manner. This chapter has fourteen exercises. Let us have a look at some of the important concepts that are discussed in this chapter.

• Definition and meaning of inverse trigonometric functions
• The inverse of the sine function
• The inverse of the cosine function
• The inverse of the tangent function
• The inverse of the secant function
• The inverse of the cosecant function
• The inverse of the cotangent function
• Properties of inverse trigonometric functions

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

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

Exercise 4.1 Page No: 4.6

1. Find the principal value of the following:

Solution:

(iii) Given functions can be written as

(iv) The given question can be written as

(v) Let

(vi) Let

2.

(i)

(ii)

Solution:

(i) The given question can be written as,

(ii) Given question can be written as

Exercise 4.2 Page No: 4.10

1. 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]

2. 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]

Exercise 4.3 Page No: 4.14

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

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 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).

Exercise 4.5 Page No: 4.21

1. 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

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

Exercise 4.7 Page No: 4.42

1. 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)

(vi) sin^-1{(sin – 17π/8)}

(vii) sin^-1(sin 3)

(viii) sin^-1(sin 4)

(ix) sin^-1(sin 12)

(x) sin^-1(sin 2)

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

(vi) Given sin^-1{(sin – 17π/8)}

But we know that – sin θ = sin (-θ)

Therefore (sin -17π/8) = – sin 17π/8

– Sin 17π/8 = – sin (2π + π/8) [since sin (2π – θ) = -sin (θ)]

It can also be written as – sin (π/8)

– Sin (π/8) = sin (-π/8) [since – sin θ = sin (-θ)]

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 -π/8) = – π/8

(vii) Given sin^-1(sin 3)

We know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

But here x = 3, which does not lie on the above range,

Therefore we know that sin (π – x) = sin (x)

Hence sin (π – 3) = sin (3) also π – 3 ∈ [-π/2, π/2]

Sin^-1(sin 3) = π – 3

(viii) Given sin^-1(sin 4)

We know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

But here x = 4, which does not lie on the above range,

Therefore we know that sin (π – x) = sin (x)

Hence sin (π – 4) = sin (4) also π – 4 ∈ [-π/2, π/2]

Sin^-1(sin 4) = π – 4

(ix) Given sin^-1(sin 12)

We know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

But here x = 12, which does not lie on the above range,

Therefore we know that sin (2nπ – x) = sin (-x)

Hence sin (2nπ – 12) = sin (-12)

Here n = 2 also 12 – 4π ∈ [-π/2, π/2]

Sin^-1(sin 12) = 12 – 4π

(x) Given sin^-1(sin 2)

We know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

But here x = 2, which does not lie on the above range,

Therefore we know that sin (π – x) = sin (x)

Hence sin (π – 2) = sin (2) also π – 2 ∈ [-π/2, π/2]

Sin^-1(sin 2) = π – 2

2. 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)

(v) cos^-1(cos 3)

(vi) cos^-1(cos 4)

(vii) cos^-1(cos 5)

(viii) cos^-1(cos 12)

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

(v) Given cos^-1(cos 3)

We know that cos^-1(cos θ) = θ if 0 ≤ θ ≤ π

Therefore by applying this in given question we get,

Cos^-1(cos 3) = 3, 3 ∈ [0, π]

(vi) Given cos^-1(cos 4)

We have cos^–1(cos x) = x if x ϵ [0, π] ≈ [0, 3.14]

And here x = 4 which does not lie in the above range.

We know that cos (2π – x) = cos(x)

Thus, cos (2π – 4) = cos (4) so 2π–4 belongs in [0, π]

Hence cos^–1(cos 4) = 2π – 4

(vii) Given cos^-1(cos 5)

We have cos^–1(cos x) = x if x ϵ [0, π] ≈ [0, 3.14]

And here x = 5 which does not lie in the above range.

We know that cos (2π – x) = cos(x)

Thus, cos (2π – 5) = cos (5) so 2π–5 belongs in [0, π]

Hence cos^–1(cos 5) = 2π – 5

(viii) Given cos^-1(cos 12)

Cos^–1(cos x) = x if x ϵ [0, π] ≈ [0, 3.14]

And here x = 12 which does not lie in the above range.

We know cos (2nπ – x) = cos (x)

Cos (2nπ – 12) = cos (12)

Here n = 2.

Also 4π – 12 belongs in [0, π]

∴ cos^–1(cos 12) = 4π – 12

3. Evaluate each of the following:

(i) tan^-1(tan π/3)

(ii) tan^-1(tan 6π/7)

(iii) tan^-1(tan 7π/6)

(iv) tan^-1(tan 9π/4)

(v) tan^-1(tan 1)

(vi) tan^-1(tan 2)

(vii) tan^-1(tan 4)

(viii) tan^-1(tan 12)

Solution:

(i) Given tan^-1(tan π/3)

As tan^-1(tan x) = x if x ϵ [-π/2, π/2]

By applying this condition in the given question we get,

Tan^-1(tan π/3) = π/3

(ii) Given tan^-1(tan 6π/7)

We know that tan 6π/7 can be written as (π – π/7)

Tan (π – π/7) = – tan π/7

We know that tan^-1(tan x) = x if x ϵ [-π/2, π/2]

Tan^-1(tan 6π/7) = – π/7

(iii) Given tan^-1(tan 7π/6)

We know that tan 7π/6 = 1/√3

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

Tan^-1 (1/√3)

Now let tan^-1 (1/√3) = y

Tan y = 1/√3

Tan (π/6) = 1/√3

The range of the principal value of tan^-1 is (-π/2, π/2) and tan (π/6) = 1/√3

Therefore tan^-1(tan 7π/6) = π/6

(iv) Given tan^-1(tan 9π/4)

We know that tan 9π/4 = 1

By substituting this value in tan^-1(tan 9π/4) we get,

Tan^-1 (1)

Now let tan^-1 (1) = y

Tan y = 1

Tan (π/4) = 1

The range of the principal value of tan^-1 is (-π/2, π/2) and tan (π/4) = 1

Therefore tan^-1(tan 9π/4) = π/4

(v) Given tan^-1(tan 1)

But we have tan^-1(tan x) = x if x ϵ [-π/2, π/2]

By substituting this condition in given question

Tan^-1(tan 1) = 1

(vi) Given tan^-1(tan 2)

As tan^-1(tan x) = x if x ϵ [-π/2, π/2]

But here x = 2 which does not belongs to above range

We also have tan (π – θ) = –tan (θ)

Therefore tan (θ – π) = tan (θ)

Tan (2 – π) = tan (2)

Now 2 – π is in the given range

Hence tan^–1 (tan 2) = 2 – π

(vii) Given tan^-1(tan 4)

As tan^-1(tan x) = x if x ϵ [-π/2, π/2]

But here x = 4 which does not belongs to above range

We also have tan (π – θ) = –tan (θ)

Therefore tan (θ – π) = tan (θ)

Tan (4 – π) = tan (4)

Now 4 – π is in the given range

Hence tan^–1 (tan 2) = 4 – π

(viii) Given tan^-1(tan 12)

As tan^-1(tan x) = x if x ϵ [-π/2, π/2]

But here x = 12 which does not belongs to above range

We know that tan (2nπ – θ) = –tan (θ)

Tan (θ – 2nπ) = tan (θ)

Here n = 2

Tan (12 – 4π) = tan (12)

Now 12 – 4π is in the given range

∴ tan^–1 (tan 12) = 12 – 4π.

Exercise 4.8 Page No: 4.54

1. Evaluate each of the following:

(i) sin (sin^-1 7/25)

(ii) Sin (cos^-1 5/13)

(iii) Sin (tan^-1 24/7)

(iv) Sin (sec^-1 17/8)

(v) Cosec (cos^-1 8/17)

(vi) Sec (sin^-1 12/13)

(vii) Tan (cos^-1 8/17)

(viii) cot (cos^-1 3/5)

(ix) Cos (tan^-1 24/7)

Solution:

(i) Given sin (sin^-1 7/25)

Now let y = sin^-1 7/25

Sin y = 7/25 where y ∈ [0, π/2]

Substituting these values in sin (sin^-1 7/25) we get

Sin (sin^-1 7/25) = 7/25

(ii) Given Sin (cos^-1 5/13)

(iii) Given Sin (tan^-1 24/7)

(iv) Given Sin (sec^-1 17/8)

(v) Given Cosec (cos^-1 8/17)

Let cos^-1(8/17) = y

cos y = 8/17 where y ∈ [0, π/2]

Now, we have to find

Cosec (cos^-1 8/17) = cosec y

We know that,

sin² θ + cos² θ = 1

sin² θ = √ (1 – cos² θ)

So,

sin y = √ (1 – cos² y)

= √ (1 – (8/17)²)

= √ (1 – 64/289)

= √ (289 – 64/289)

= √ (225/289)

= 15/17

Hence,

Cosec y = 1/sin y = 1/ (15/17) = 17/15

Therefore,

Cosec (cos^-1 8/17) = 17/15

(vi) Given Sec (sin^-1 12/13)

(vii) Given Tan (cos^-1 8/17)

(viii) Given cot (cos^-1 3/5)

(ix) Given Cos (tan^-1 24/7)

.

Exercise 4.9 Page No: 4.58

1. Evaluate:

(i) Cos {sin^-1 (-7/25)}

(ii) Sec {cot^-1 (-5/12)}

(iii) Cot {sec^-1 (-13/5)}

Solution:

(i) Given Cos {sin^-1 (-7/25)}

(ii) Given Sec {cot^-1 (-5/12)}

(iii) Given Cot {sec^-1 (-13/5)}

Exercise 4.10 Page No: 4.66

1. Evaluate:

(i) Cot (sin^-1 (3/4) + sec^-1 (4/3))

(ii) Sin (tan^-1 x + tan^-1 1/x) for x < 0

(iii) Sin (tan^-1 x + tan^-1 1/x) for x > 0

(iv) Cot (tan^-1 a + cot^-1 a)

(v) Cos (sec^-1 x + cosec^-1 x), |x| ≥ 1

Solution:

(i) Given Cot (sin^-1 (3/4) + sec^-1 (4/3))

(ii) Given Sin (tan^-1 x + tan^-1 1/x) for x < 0

(iii) Given Sin (tan^-1 x + tan^-1 1/x) for x > 0

(iv) Given Cot (tan^-1 a + cot^-1 a)

(v) Given Cos (sec^-1 x + cosec^-1 x), |x| ≥ 1

= 0

2. If cos^-1 x + cos^-1 y = π/4, find the value of sin^-1 x + sin^-1 y.

Solution:

Given cos^-1 x + cos^-1 y = π/4

3. If sin^-1 x + sin^-1 y = π/3 and cos^-1 x – cos^-1 y = π/6, find the values of x and y.

Solution:

Given sin^-1 x + sin^-1 y = π/3 ……. Equation (i)

And cos^-1 x – cos^-1 y = π/6 ……… Equation (ii)

4. If cot (cos^-1 3/5 + sin^-1 x) = 0, find the value of x.

Solution:

Given cot (cos^-1 3/5 + sin^-1 x) = 0

On rearranging, we get,

(cos^-1 3/5 + sin^-1 x) = cot^-1 (0)

(Cos^-1 3/5 + sin^-1 x) = π/2

We know that cos^-1 x + sin^-1 x = π/2

Then sin^-1 x = π/2 – cos^-1 x

Substituting the above in (cos^-1 3/5 + sin^-1 x) = π/2 we get,

(Cos^-1 3/5 + π/2 – cos^-1 x) = π/2

Now on rearranging, we get,

(Cos^-1 3/5 – cos^-1 x) = π/2 – π/2

(Cos^-1 3/5 – cos^-1 x) = 0

Therefore Cos^-1 3/5 = cos^-1 x

On comparing the above equation, we get,

x = 3/5

5. If (sin^-1 x)² + (cos^-1 x)² = 17 π²/36, find x.

Solution:

Given (sin^-1 x)² + (cos^-1 x)² = 17 π²/36

We know that cos^-1 x + sin^-1 x = π/2

Then cos^-1 x = π/2 – sin^-1 x

Substituting this in (sin^-1 x)² + (cos^-1 x)² = 17 π²/36 we get

(sin^-1 x)² + (π/2 – sin^-1 x)² = 17 π²/36

Let y = sin^-1 x

y² + ((π/2) – y)² = 17 π²/36

y² + π²/4 – y² – 2y ((π/2) – y) = 17 π²/36

π²/4 – πy + 2 y² = 17 π²/36

On rearranging and simplifying, we get

2y² – πy + 2/9 π² = 0

18y² – 9 πy + 2 π² = 0

18y² – 12 πy + 3 πy + 2 π² = 0

6y (3y – 2π) + π (3y – 2π) = 0

Now, (3y – 2π) = 0 and (6y + π) = 0

Therefore y = 2π/3 and y = – π/6

Now substituting y = – π/6 in y = sin^-1 x we get

sin^-1 x = – π/6

x = sin (- π/6)

x = -1/2

Now substituting y = -2π/3 in y = sin^-1 x we get

x = sin (2π/3)

x = √3/2

Now substituting x = √3/2 in (sin^-1 x)² + (cos^-1 x)² = 17 π²/36 we get,

= π/3 + π/6

= π/2 which is not equal to 17 π²/36

So we have to neglect this root.

Now substituting x = -1/2 in (sin^-1 x)² + (cos^-1 x)² = 17 π²/36 we get,

= π²/36 + 4 π²/9

= 17 π²/36

Hence x = -1/2.

Exercise 4.11 Page No: 4.82

1. Prove the following results:

(i) Tan^-1 (1/7) + tan^-1 (1/13) = tan^-1 (2/9)

(ii) Sin^-1 (12/13) + cos^-1 (4/5) + tan^-1 (63/16) = π

(iii) tan^-1 (1/4) + tan^-1 (2/9) = Sin^-1 (1/ √5)

Solution:

(i) Given Tan^-1 (1/7) + tan^-1 (1/13) = tan^-1 (2/9)

Hence, proved.

(ii) Given Sin^-1 (12/13) + cos^-1 (4/5) + tan^-1 (63/16) = π

Consider LHS

Hence, proved.

(iii) Given tan^-1 (1/4) + tan^-1 (2/9) = Sin^-1 (1/ √5)

2. Find the value of tan^-1 (x/y) – tan^-1 {(x-y)/(x + y)}

Solution:

Given tan^-1 (x/y) – tan^-1 {(x-y)/(x + y)}

Exercise 4.12 Page No: 4.89

1. Evaluate: Cos (sin ^-1 3/5 + sin^-1 5/13)

Solution:

Given Cos (sin ^-1 3/5 + sin^-1 5/13)

We know that,

Exercise 4.13 Page No: 4.92

1. If cos^-1 (x/2) + cos^-1 (y/3) = α, then prove that 9x² – 12xy cos α + 4y² = 36 sin² α

Solution:

Given cos^-1 (x/2) + cos^-1 (y/3) = α

Hence, proved.

2. Solve the equation: cos^-1 (a/x) – cos^-1 (b/x) = cos^-1 (1/b) – cos^-1 (1/a)

Solution:

Given cos^-1 (a/x) – cos^-1 (b/x) = cos^-1 (1/b) – cos^-1 (1/a)

Exercise 4.14 Page No: 4.115

1. Evaluate the following:

(i) tan {2 tan^-1 (1/5) – π/4}

(ii) Tan {1/2 sin^-1 (3/4)}

(iii) Sin {1/2 cos^-1 (4/5)}

(iv) Sin (2 tan ^-1 2/3) + cos (tan^-1 √3)

Solution:

(i) Given tan {2 tan^-1 (1/5) – π/4}

(ii) Given tan {1/2 sin^-1 (3/4)}

(iii) Given sin {1/2 cos^-1 (4/5)}

(iv) Given Sin (2 tan ^-1 2/3) + cos (tan^-1 √3)

2. Prove the following results:

(i) 2 sin^-1 (3/5) = tan^-1 (24/7)

(ii) tan^-1 ¼ + tan^-1 (2/9) = ½ cos^-1 (3/5) = ½ sin^-1 (4/5)

(iii) tan^-1 (2/3) = ½ tan^-1 (12/5)

(iv) tan^-1 (1/7) + 2 tan^-1 (1/3) = π/4

(v) sin^-1 (4/5) + 2 tan^-1 (1/3) = π/2

(vi) 2 sin^-1 (3/5) – tan^-1 (17/31) = π/4

(vii) 2 tan^-1 (1/5) + tan^-1 (1/8) = tan^-1 (4/7)

(viii) 2 tan^-1 (3/4) – tan^-1 (17/31) = π/4

(ix) 2 tan^-1 (1/2) + tan^-1 (1/7) = tan^-1 (31/17)

(x) 4 tan^-1(1/5) – tan^-1(1/239) = π/4

Solution:

(i) Given 2 sin^-1 (3/5) = tan^-1 (24/7)

Hence, proved.

(ii) Given tan^-1 ¼ + tan^-1 (2/9) = ½ cos^-1 (3/5) = ½ sin^-1 (4/5)

Hence, proved.

(iii) Given tan^-1 (2/3) = ½ tan^-1 (12/5)

Hence, proved.

(iv) Given tan^-1 (1/7) + 2 tan^-1 (1/3) = π/4

Hence, proved.

(v) Given sin^-1 (4/5) + 2 tan^-1 (1/3) = π/2

(vi) Given 2 sin^-1 (3/5) – tan^-1 (17/31) = π/4

(vii) Given 2 tan^-1 (1/5) + tan^-1 (1/8) = tan^-1 (4/7)

Hence, proved.

(viii) Given 2 tan^-1 (3/4) – tan^-1 (17/31) = π/4

Hence, proved.

(ix) Given 2 tan^-1 (1/2) + tan^-1 (1/7) = tan^-1 (31/17)

Hence, proved.

(x) Given 4 tan^-1(1/5) – tan^-1(1/239) = π/4

Hence, proved.

3. If sin^-1 (2a/1 + a²) – cos^-1(1 – b²/1 + b²) = tan^-1(2x/1 – x²), then prove that x = (a – b)/ (1 + a b)

Solution:

Given sin^-1 (2a/1 + a²) – cos^-1(1 – b²/1 + b²) = tan^-1(2x/1 – x²)

Hence, proved.

4. Prove that:

(i) tan^-1{(1 – x²)/ 2x)} + cot^-1{(1 – x²)/ 2x)} = π/2

(ii) sin {tan^-1 (1 – x²)/ 2x) + cos^-1 (1 – x²)/ (1 + x²)} = 1

Solution:

(i) Given tan^-1{(1 – x²)/ 2x)} + cot^-1{(1 – x²)/ 2x)} = π/2

Hence, proved.

(ii) Given sin {tan^-1 (1 – x²)/ 2x) + cos^-1 (1 – x²)/ (1 + x²)}

Hence, proved.

5. If sin^-1 (2a/ 1+ a²) + sin^-1 (2b/ 1+ b²) = 2 tan^-1 x, prove that x = (a + b/ 1 – a b)

Solution:

Given sin^-1 (2a/ 1+ a²) + sin^-1 (2b/ 1+ b²) = 2 tan^-1 x

Hence, proved.

Also, access exercises of RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions

Exercise 4.1 Solutions

Exercise 4.2 Solutions

Exercise 4.3 Solutions

Exercise 4.4 Solutions

Exercise 4.5 Solutions

Exercise 4.6 Solutions

Exercise 4.7 Solutions

Exercise 4.8 Solutions

Exercise 4.9 Solutions

Exercise 4.10 Solutions

Exercise 4.11 Solutions

Exercise 4.12 Solutions

Exercise 4.13 Solutions

Exercise 4.14 Solutions