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

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, which match the grasping abilities of students. To make this possible, our set of faculty has provided exercise-wise solutions for each chapter as per 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 in a stepwise manner as per the exam pattern and mark weightage. RD Sharma Solutions for Class 12 Chapter 4 Inverse Trigonometric Functions PDF are given here. 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
  • Inverse of sine function
  • Inverse of cosine function
  • Inverse of tangent function
  • Inverse of secant function
  • Inverse of cosecant function
  • Inverse of cotangent function
  • Properties of inverse trigonometric functions

Download the PDF of 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:

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 1

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 2

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 3

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 5

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 6

Solution:

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(iii) Given functions can be written as

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 9

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 10

(iv) The given question can be written as

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 11

(v) Let

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

(vi) Let

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 13

2.

(i) RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 14

(ii) RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 15

Solution:

(i) The given question can be written as,

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 16

(ii) Given question can be written as

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 17



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)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 18

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 19

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 20

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 21

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 22

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 23

(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)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 26

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 27

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 28

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 29

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 30

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 31

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 32

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 33

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 34

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 35.

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 36


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)}

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 37

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 38

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 39

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 40

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 41

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 42


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))

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 43

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 44

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 45

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 46

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 47

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 48

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 49

= 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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 50

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)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 51

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 52

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 53

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 54

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)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 55

Hence, proved.

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

Consider LHS

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 56

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 57

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 58

Hence, proved.

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 59

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 60

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)}

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 61


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,

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 62

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 62a


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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 63

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 64

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)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 65

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 66

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 67


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}

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 68

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 69

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 70

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 71

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 72

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 73

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 74

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 75

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 76

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)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 77

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 78

Hence, proved.

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 79

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 80

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 81

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 82

Hence, proved.

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 83

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 84

Hence, proved.

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 85

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 86

Hence, proved.

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 87

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 88

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 89

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 90

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 91

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 92

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 93

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 94

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 95

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 96

Hence, proved.

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 97

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 98

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 99

Hence, proved.

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 100

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 101

Hence, proved.

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 102

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 103

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)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 104

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 105

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 106

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 107

Hence, proved.

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 108

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 109

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 110

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 111

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 112
RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 113

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

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