# 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 (2021-22)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions is provided in a comprehensive manner for students to secure good marks in their academics. Class 12 is a crucial stage in a studentâ€™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 learning fun and interesting, 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.Â

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. 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
• 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

### 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 â€“ 4)

Solution:

Given f(x) = cos -1 (x2 â€“ 4)

We know that domain of cos-1 (x2 â€“ 4) lies in the interval [-1, 1]

Therefore, we can write as

-1 â‰¤ x2 â€“ 4 â‰¤ 1

4 – 1 â‰¤ x2 â‰¤ 1 + 4

3 â‰¤ x2 â‰¤ 5

Â±âˆšÂ 3 â‰¤ x â‰¤ Â±âˆš5

– âˆš5 â‰¤ x â‰¤ – âˆš3 and âˆš3 â‰¤ x â‰¤ âˆš5

Therefore domain of cos-1 (x2 â€“ 4) is [- âˆš5, – âˆš3] âˆª [âˆš3, âˆš5]

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 â‰¤ 2x â‰¤ 1

– Â½ â‰¤ x â‰¤ Â½

Hence, domain of cos-1(2x) + sin-1 x lies in the interval [- Â½, Â½]

### Exercise 4.3 Page No: 4.14

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

(i) tan-1 (1/âˆš3)

(ii) tan-1 (-1/âˆš3)

(iii) tan-1 (cos (Ï€/2))

(iv) tan-1 (2 cos (2Ï€/3))

Solution:

(i) Given tan-1 (1/âˆš3)

We know that for any x âˆˆ R, tan-1 represents an angle in (-Ï€/2, Ï€/2) whose tangent is x.

So, tan-1 (1/âˆš3) = an angle in (-Ï€/2, Ï€/2) whose tangent is (1/âˆš3)

But we know that the value is equal to Ï€/6

Therefore tan-1 (1/âˆš3) = Ï€/6

Hence the principal value of tan-1 (1/âˆš3) = Ï€/6

(ii) Given tan-1 (-1/âˆš3)

We know that for any x âˆˆ R, tan-1 represents an angle in (-Ï€/2, Ï€/2) whose tangent is x.

So, tan-1 (-1/âˆš3) = an angle in (-Ï€/2, Ï€/2) whose tangent is (1/âˆš3)

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

Therefore tan-1 (-1/âˆš3) = -Ï€/6

Hence the principal value of tan-1 (-1/âˆš3) = – Ï€/6

(iii) Given that tan-1 (cos (Ï€/2))

But we know that cos (Ï€/2) = 0

We know that for any x âˆˆ R, tan-1 represents an angle in (-Ï€/2, Ï€/2) whose tangent is x.

Therefore tan-1 (0) = 0

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

(iv) Given that tan-1 (2 cos (2Ï€/3))

But we know that cos Ï€/3 = 1/2

So, cos (2Ï€/3) = -1/2

Therefore tan-1 (2 cos (2Ï€/3)) = tan-1 (2 Ã— – Â½)

= tan-1(-1)

= – Ï€/4

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

### Exercise 4.4 Page No: 4.18

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

(i) sec-1 (-âˆš2)

(ii) sec-1 (2)

(iii) sec-1 (2 sin (3Ï€/4))

(iv) sec-1 (2 tan (3Ï€/4))

Solution:

(i) Given sec-1 (-âˆš2)

Now let y = sec-1 (-âˆš2)

Sec y = -âˆš2

We know that sec Ï€/4 = âˆš2

Therefore, -sec (Ï€/4) = -âˆš2

= sec (Ï€ – Ï€/4)

= sec (3Ï€/4)

Thus the range of principal value of sec-1 is [0, Ï€] â€“ {Ï€/2}

And sec (3Ï€/4) = – âˆš2

Hence the principal value of sec-1 (-âˆš2) is 3Ï€/4

(ii) Given sec-1 (2)

Let y = sec-1 (2)

Sec y = 2

= Sec Ï€/3

Therefore the range of principal value of sec-1 is [0, Ï€] â€“ {Ï€/2} and sec Ï€/3 = 2

Thus the principal value of sec-1 (2) is Ï€/3

(iii) Given sec-1 (2 sin (3Ï€/4))

But we know that sin (3Ï€/4) = 1/âˆš2

Therefore 2 sin (3Ï€/4) = 2 Ã— 1/âˆš2

2 sin (3Ï€/4) = âˆš2

Therefore by substituting above values in sec-1 (2 sin (3Ï€/4)), we get

Sec-1 (âˆš2)

Let Sec-1 (âˆš2) = y

Sec y = âˆš2

Sec (Ï€/4) = âˆš2

Therefore range of principal value of sec-1 is [0, Ï€] â€“ {Ï€/2} and sec (Ï€/4) = âˆš2

Thus the principal value of sec-1 (2 sin (3Ï€/4)) is Ï€/4.

(iv) Given sec-1 (2 tan (3Ï€/4))

But we know that tan (3Ï€/4) = -1

Therefore, 2 tan (3Ï€/4) = 2 Ã— -1

2 tan (3Ï€/4) = -2

By substituting these values in sec-1 (2 tan (3Ï€/4)), we get

Sec-1 (-2)

Now let y = Sec-1 (-2)

Sec y = – 2

– sec (Ï€/3) = -2

= sec (Ï€ â€“ Ï€/3)

= sec (2Ï€/3)

Therefore the range of principal value of sec-1 is [0, Ï€] â€“ {Ï€/2} and sec (2Ï€/3) = -2

Thus, the principal value of sec-1 (2 tan (3Ï€/4)) is (2Ï€/3).

### Exercise 4.5 Page No: 4.21

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

(i) cosec-1 (-âˆš2)

(ii) cosec-1 (-2)

(iii) cosec-1 (2/âˆš3)

(iv) cosec-1 (2 cos (2Ï€/3))

Solution:

(i) Given cosec-1 (-âˆš2)

Let y = cosec-1 (-âˆš2)

Cosec y = -âˆš2

– Cosec y = âˆš2

– Cosec (Ï€/4) = âˆš2

– Cosec (Ï€/4) = cosec (-Ï€/4) [since â€“cosec Î¸ = cosec (-Î¸)]

The range of principal value of cosec-1 [-Ï€/2, Ï€/2] â€“ {0} and cosec (-Ï€/4) = – âˆš2

Cosec (-Ï€/4) = – âˆš2

Therefore the principal value of cosec-1 (-âˆš2) is – Ï€/4

(ii) Given cosec-1 (-2)

Let y = cosec-1 (-2)

Cosec y = -2

– Cosec y = 2

– Cosec (Ï€/6) = 2

– Cosec (Ï€/6) = cosec (-Ï€/6) [since â€“cosec Î¸ = cosec (-Î¸)]

The range of principal value of cosec-1 [-Ï€/2, Ï€/2] â€“ {0} and cosec (-Ï€/6) = – 2

Cosec (-Ï€/6) = – 2

Therefore the principal value of cosec-1 (-2) is – Ï€/6

(iii) Given cosec-1 (2/âˆš3)

Let y = cosec-1 (2/âˆš3)

Cosec y = (2/âˆš3)

Cosec (Ï€/3) = (2/âˆš3)

Therefore range of principal value of cosec-1 is [-Ï€/2, Ï€/2] â€“ {0} and cosec (Ï€/3) = (2/âˆš3)

Thus, the principal value of cosec-1 (2/âˆš3) is Ï€/3

(iv) Given cosec-1 (2 cos (2Ï€/3))

But we know that cos (2Ï€/3) = – Â½

Therefore 2 cos (2Ï€/3) = 2 Ã— – Â½

2 cos (2Ï€/3) = -1

By substituting these values in cosec-1 (2 cos (2Ï€/3)) we get,

Cosec-1 (-1)

Let y = cosec-1 (-1)

– Cosec y = 1

– Cosec (Ï€/2) = cosec (-Ï€/2) [since â€“cosec Î¸ = cosec (-Î¸)]

The range of principal value of cosec-1 [-Ï€/2, Ï€/2] â€“ {0} and cosec (-Ï€/2) = – 1

Cosec (-Ï€/2) = – 1

Therefore the principal value of cosec-1 (2 cos (2Ï€/3)) is – Ï€/2

### Exercise 4.6 Page No: 4.24

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

(i) cot-1(-âˆš3)

(ii) Cot-1(âˆš3)

(iii) cot-1(-1/âˆš3)

(iv) cot-1(tan 3Ï€/4)

Solution:

(i) Given cot-1(-âˆš3)

Let y = cot-1(-âˆš3)

– Cot (Ï€/6) = âˆš3

= Cot (Ï€ â€“ Ï€/6)

= cot (5Ï€/6)

The range of principal value of cot-1 is (0, Ï€) and cot (5 Ï€/6) = – âˆš3

Thus, the principal value of cot-1 (- âˆš3) is 5Ï€/6

(ii) Given Cot-1(âˆš3)

Let y = cot-1(âˆš3)

Cot (Ï€/6) = âˆš3

The range of principal value of cot-1 is (0, Ï€) and

Thus, the principal value of cot-1 (âˆš3) is Ï€/6

(iii) Given cot-1(-1/âˆš3)

Let y = cot-1(-1/âˆš3)

Cot y = (-1/âˆš3)

– Cot (Ï€/3) = 1/âˆš3

= Cot (Ï€ â€“ Ï€/3)

= cot (2Ï€/3)

The range of principal value of cot-1(0, Ï€) and cot (2Ï€/3) = – 1/âˆš3

Therefore the principal value of cot-1(-1/âˆš3) is 2Ï€/3

(iv) Given cot-1(tan 3Ï€/4)

But we know that tan 3Ï€/4 = -1

By substituting this value in cot-1(tan 3Ï€/4) we get

Cot-1(-1)

Now, let y = cot-1(-1)

Cot y = (-1)

– Cot (Ï€/4) = 1

= Cot (Ï€ â€“ Ï€/4)

= cot (3Ï€/4)

The range of principal value of cot-1(0, Ï€) and cot (3Ï€/4) = – 1

Therefore the principal value of cot-1(tan 3Ï€/4) is 3Ï€/4

### Exercise 4.7 Page No: 4.42

1. Evaluate each of the following:

(i) sin-1(sin Ï€/6)

(ii) sin-1(sin 7Ï€/6)

(iii) sin-1(sin 5Ï€/6)

(iv) sin-1(sin 13Ï€/7)

(v) sin-1(sin 17Ï€/8)

(vi) sin-1{(sin – 17Ï€/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 Ï€/6)

We know that the value of sin Ï€/6 is Â½

By substituting this value in sin-1(sin Ï€/6)

We get, sin-1 (1/2)

Now let y = sin-1 (1/2)

Sin (Ï€/6) = Â½

The range of principal value of sin-1(-Ï€/2, Ï€/2) and sin (Ï€/6) = Â½

Therefore sin-1(sin Ï€/6) = Ï€/6

(ii) Given sin-1(sin 7Ï€/6)

But we know that sin 7Ï€/6 = – Â½

By substituting this in sin-1(sin 7Ï€/6) we get,

Sin-1 (-1/2)

Now let y = sin-1 (-1/2)

– Sin y = Â½

– Sin (Ï€/6) = Â½

– Sin (Ï€/6) = sin (- Ï€/6)

The range of principal value of sin-1(-Ï€/2, Ï€/2) and sin (- Ï€/6) = – Â½

Therefore sin-1(sin 7Ï€/6) = – Ï€/6

(iii) Given sin-1(sin 5Ï€/6)

We know that the value of sin 5Ï€/6 is Â½

By substituting this value in sin-1(sin 5Ï€/6)

We get, sin-1 (1/2)

Now let y = sin-1 (1/2)

Sin (Ï€/6) = Â½

The range of principal value of sin-1(-Ï€/2, Ï€/2) and sin (Ï€/6) = Â½

Therefore sin-1(sin 5Ï€/6) = Ï€/6

(iv) Given sin-1(sin 13Ï€/7)

Given question can be written as sin (2Ï€ â€“ Ï€/7)

Sin (2Ï€ â€“ Ï€/7) can be written as sin (-Ï€/7) [since sin (2Ï€ â€“ Î¸) = sin (-Î¸)]

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

As sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/2]

Therefore sin-1(sin 13Ï€/7) = – Ï€/7

(v) Given sin-1(sin 17Ï€/8)

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

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

By substituting these values in sin-1(sin 17Ï€/8) we get sin-1(sin Ï€/8)

As sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/2]

Therefore sin-1(sin 17Ï€/8) = Ï€/8

(vi) Given sin-1{(sin – 17Ï€/8)}

But we know that â€“ sin Î¸ = sin (-Î¸)

Therefore (sin -17Ï€/8) = – sin 17Ï€/8

– Sin 17Ï€/8 = – sin (2Ï€ + Ï€/8) [since sin (2Ï€ â€“ Î¸) = -sin (Î¸)]

It can also be written as â€“ sin (Ï€/8)

â€“ Sin (Ï€/8) = sin (-Ï€/8) [since â€“ sin Î¸ = sin (-Î¸)]

By substituting these values in sin-1{(sin – 17Ï€/8)} we get,

Sin-1(sin – Ï€/8)

As sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/2]

Therefore sin-1(sin -Ï€/8) = – Ï€/8

(vii) Given sin-1(sin 3)

We know that sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/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 (Ï€ â€“ x) = sin (x)

Hence sin (Ï€ â€“ 3) = sin (3) also Ï€ â€“ 3 âˆˆ [-Ï€/2, Ï€/2]

Sin-1(sin 3) = Ï€ â€“ 3

(viii) Given sin-1(sin 4)

We know that sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/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 (Ï€ â€“ x) = sin (x)

Hence sin (Ï€ â€“ 4) = sin (4) also Ï€ â€“ 4 âˆˆ [-Ï€/2, Ï€/2]

Sin-1(sin 4) = Ï€ â€“ 4

(ix) Given sin-1(sin 12)

We know that sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/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 (2nÏ€ â€“ x) = sin (-x)

Hence sin (2nÏ€ â€“ 12) = sin (-12)

Here n = 2 also 12 â€“ 4Ï€ âˆˆ [-Ï€/2, Ï€/2]

Sin-1(sin 12) = 12 â€“ 4Ï€

(x) Given sin-1(sin 2)

We know that sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/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 (Ï€ â€“ x) = sin (x)

Hence sin (Ï€ â€“ 2) = sin (2) also Ï€ â€“ 2 âˆˆ [-Ï€/2, Ï€/2]

Sin-1(sin 2) = Ï€ â€“ 2

2. Evaluate each of the following:

(i) cos-1{cos (-Ï€/4)}

(ii) cos-1(cos 5Ï€/4)

(iii) cos-1(cos 4Ï€/3)

(iv) cos-1(cos 13Ï€/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 (-Ï€/4)}

We know that cos (-Ï€/4) = cos (Ï€/4) [since cos (-Î¸) = cos Î¸

Also know that cos (Ï€/4) = 1/âˆš2

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

Cos-1(1/âˆš2)

Now let y = cos-1(1/âˆš2)

Therefore cos y = 1/âˆš2

Hence range of principal value of cos-1 is [0, Ï€] and cos (Ï€/4) = 1/âˆš2

Therefore cos-1{cos (-Ï€/4)} = Ï€/4

(ii) Given cos-1(cos 5Ï€/4)

But we know that cos (5Ï€/4) = -1/âˆš2

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

Cos-1(-1/âˆš2)

Now let y = cos-1(-1/âˆš2)

Therefore cos y = – 1/âˆš2

– Cos (Ï€/4) = 1/âˆš2

Cos (Ï€ – Ï€/4) = – 1/âˆš2

Cos (3 Ï€/4) = – 1/âˆš2

Hence range of principal value of cos-1 is [0, Ï€] and cos (3Ï€/4) = -1/âˆš2

Therefore cos-1{cos (5Ï€/4)} = 3Ï€/4

(iii) Given cos-1(cos 4Ï€/3)

But we know that cos (4Ï€/3) = -1/2

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

Cos-1(-1/2)

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

Therefore cos y = – 1/2

– Cos (Ï€/3) = 1/2

Cos (Ï€ – Ï€/3) = – 1/2

Cos (2Ï€/3) = – 1/2

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

Therefore cos-1{cos (4Ï€/3)} = 2Ï€/3

(iv) Given cos-1(cos 13Ï€/6)

But we know that cos (13Ï€/6) = âˆš3/2

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

Cos-1(âˆš3/2)

Now let y = cos-1(âˆš3/2)

Therefore cos y = âˆš3/2

Cos (Ï€/6) = âˆš3/2

Hence range of principal value of cos-1 is [0, Ï€] and cos (Ï€/6) = âˆš3/2

Therefore cos-1{cos (13Ï€/6)} = Ï€/6

(v) Given cos-1(cos 3)

We know that cos-1(cos Î¸) = Î¸ if 0 â‰¤ Î¸ â‰¤ Ï€

Therefore by applying this in given question we get,

Cos-1(cos 3) = 3, 3 âˆˆ [0, Ï€]

(vi) Given cos-1(cos 4)

We have cosâ€“1(cos x) = x if xÂ ÏµÂ [0, Ï€] â‰ˆ [0, 3.14]

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

We know that cos (2Ï€ â€“ x) = cos(x)

Thus, cos (2Ï€ â€“ 4) = cos (4) so 2Ï€â€“4 belongs in [0, Ï€]

Hence cosâ€“1(cos 4) = 2Ï€ â€“ 4

(vii) Given cos-1(cos 5)

We have cosâ€“1(cos x) = x if xÂ ÏµÂ [0, Ï€] â‰ˆ [0, 3.14]

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

We know that cos (2Ï€ â€“ x) = cos(x)

Thus, cos (2Ï€ â€“ 5) = cos (5) so 2Ï€â€“5 belongs in [0, Ï€]

Hence cosâ€“1(cos 5) = 2Ï€ â€“ 5

(viii) Given cos-1(cos 12)

Cosâ€“1(cos x) = x if xÂ ÏµÂ [0, Ï€] â‰ˆ [0, 3.14]

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

We know cos (2nÏ€ â€“ x) = cos (x)

Cos (2nÏ€ â€“ 12) = cos (12)

Here n = 2.

Also 4Ï€ â€“ 12 belongs in [0, Ï€]

âˆ´Â cosâ€“1(cos 12) = 4Ï€ â€“ 12

3. Evaluate each of the following:

(i) tan-1(tan Ï€/3)

(ii) tan-1(tan 6Ï€/7)

(iii) tan-1(tan 7Ï€/6)

(iv) tan-1(tan 9Ï€/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 Ï€/3)

As tan-1(tan x) = x if x Ïµ [-Ï€/2, Ï€/2]

By applying this condition in the given question we get,

Tan-1(tan Ï€/3) = Ï€/3

(ii) Given tan-1(tan 6Ï€/7)

We know that tan 6Ï€/7 can be written as (Ï€ â€“ Ï€/7)

Tan (Ï€ â€“ Ï€/7) = – tan Ï€/7

We know that tan-1(tan x) = x if x Ïµ [-Ï€/2, Ï€/2]

Tan-1(tan 6Ï€/7) = – Ï€/7

(iii) Given tan-1(tan 7Ï€/6)

We know that tan 7Ï€/6 = 1/âˆš3

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

Tan-1 (1/âˆš3)

Now let tan-1 (1/âˆš3) = y

Tan y = 1/âˆš3

Tan (Ï€/6) = 1/âˆš3

The range of the principal value of tan-1 is (-Ï€/2, Ï€/2) and tan (Ï€/6) = 1/âˆš3

Therefore tan-1(tan 7Ï€/6) = Ï€/6

(iv) Given tan-1(tan 9Ï€/4)

We know that tan 9Ï€/4 = 1

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

Tan-1 (1)

Now let tan-1 (1) = y

Tan y = 1

Tan (Ï€/4) = 1

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

Therefore tan-1(tan 9Ï€/4) = Ï€/4

(v) Given tan-1(tan 1)

But we have tan-1(tan x) = x if x Ïµ [-Ï€/2, Ï€/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 Ïµ [-Ï€/2, Ï€/2]

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

We also have tan (Ï€ â€“ Î¸) = â€“tan (Î¸)

Therefore tan (Î¸ â€“ Ï€) = tan (Î¸)

Tan (2 â€“ Ï€) = tan (2)

Now 2 â€“ Ï€ is in the given range

Hence tanâ€“1Â (tan 2) = 2 â€“ Ï€

(vii) Given tan-1(tan 4)

As tan-1(tan x) = x if x Ïµ [-Ï€/2, Ï€/2]

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

We also have tan (Ï€ â€“ Î¸) = â€“tan (Î¸)

Therefore tan (Î¸ â€“ Ï€) = tan (Î¸)

Tan (4 â€“ Ï€) = tan (4)

Now 4 â€“ Ï€ is in the given range

Hence tanâ€“1Â (tan 2) = 4 â€“ Ï€

(viii) Given tan-1(tan 12)

As tan-1(tan x) = x if x Ïµ [-Ï€/2, Ï€/2]

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

We know that tan (2nÏ€ â€“ Î¸) = â€“tan (Î¸)

Tan (Î¸ â€“ 2nÏ€) = tan (Î¸)

Here n = 2

Tan (12 â€“ 4Ï€) = tan (12)

Now 12 â€“ 4Ï€ is in the given range

âˆ´Â tanâ€“1Â (tan 12) = 12 â€“ 4Ï€.

### 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)

(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)

Solution:

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

Now let y = sin-1 7/25

Sin y = 7/25 where y âˆˆ [0, Ï€/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)

(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 âˆˆ [0, Ï€/2]

Now, we have to find

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

We know that,

sin2 Î¸ + cos2 Î¸ = 1

sin2 Î¸ = âˆš (1 â€“ cos2 Î¸)

So,

sin y = âˆš (1 â€“ cos2 y)

= âˆš (1 â€“ (8/17)2)

= âˆš (1 â€“ 64/289)

= âˆš (289 â€“ 64/289)

= âˆš (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)

.

### 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| â‰¥ 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| â‰¥ 1

= 0

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

Solution:

Given cos-1 x + cos-1 y = Ï€/4

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

Solution:

Given sin-1 x + sin-1 y = Ï€/3 â€¦â€¦. Equation (i)

And cos-1 x – cos-1 y = Ï€/6 â€¦â€¦â€¦ 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) = Ï€/2

We know that cos-1 x + sin-1 x = Ï€/2

Then sin-1 x = Ï€/2 – cos-1 x

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

(Cos-1 3/5 + Ï€/2 – cos-1 x) = Ï€/2

Now on rearranging we get,

(Cos-1 3/5 – cos-1 x) = Ï€/2 – Ï€/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 Ï€2/36, find x.

Solution:

Given (sin-1 x)2 + (cos-1 x)2 = 17 Ï€2/36

We know that cos-1 x + sin-1 x = Ï€/2

Then cos-1 x = Ï€/2 – sin-1 x

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

(sin-1 x)2 + (Ï€/2 – sin-1 x)2 = 17 Ï€2/36

Let y = sin-1 x

y2 + ((Ï€/2) â€“ y)2 = 17 Ï€2/36

y2 + Ï€2/4 â€“ y2 â€“ 2y ((Ï€/2) â€“ y) = 17 Ï€2/36

Ï€2/4 â€“ Ï€y + 2 y2 = 17 Ï€2/36

On rearranging and simplifying, we get

2y2 – Ï€y + 2/9 Ï€2 = 0

18y2 – 9 Ï€y + 2 Ï€2 = 0

18y2 – 12 Ï€y + 3 Ï€y + 2 Ï€2 = 0

6y (3y – 2Ï€) + Ï€ (3y – 2Ï€) = 0

Now, (3y – 2Ï€) = 0 and (6y + Ï€) = 0

Therefore y = 2Ï€/3 and y = – Ï€/6

Now substituting y = – Ï€/6 in y = sin-1 x we get

sin-1 x = – Ï€/6

x = sin (- Ï€/6)

x = -1/2

Now substituting y = -2Ï€/3 in y = sin-1 x we get

x = sin (2Ï€/3)

x = âˆš3/2

Now substituting x = âˆš3/2 in (sin-1 x)2 + (cos-1 x)2 = 17 Ï€2/36 we get,

= Ï€/3 + Ï€/6

= Ï€/2 which is not equal to 17 Ï€2/36

So we have to neglect this root.

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

= Ï€2/36 + 4 Ï€2/9

= 17 Ï€2/36

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) = Ï€

(iii) tan-1 (1/4) + tan-1 (2/9) = Sin-1 (1/ âˆš5)

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) = Ï€

Consider LHS

Hence, proved.

(iii) Given tan-1 (1/4) + tan-1 (2/9) = Sin-1 (1/ âˆš5)

2. Find the value of tan-1 (x/y) â€“ tan-1 {(x-y)/(x + y)}

Solution:

Given tan-1 (x/y) â€“ 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) = Î±, then prove that 9x2 â€“ 12xy cos Î± + 4y2 = 36 sin2 Î±

Solution:

Given cos-1 (x/2) + cos-1 (y/3) = Î±

Hence, proved.

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

Solution:

Given cos-1 (a/x) â€“ 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) â€“ Ï€/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 âˆš3)

Solution:

(i) Given tan {2 tan-1 (1/5) â€“ Ï€/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 âˆš3)

2. Prove the following results:

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

(ii) tan-1 Â¼ + tan-1 (2/9) = Â½ cos-1 (3/5) = Â½ sin-1 (4/5)

(iii) tan-1 (2/3) = Â½ tan-1 (12/5)

(iv) tan-1 (1/7) + 2 tan-1 (1/3) = Ï€/4

(v) sin-1 (4/5) + 2 tan-1 (1/3) = Ï€/2

(vi) 2 sin-1 (3/5) â€“ tan-1 (17/31) = Ï€/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) = Ï€/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) = Ï€/4

Solution:

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

Hence, proved.

(ii) Given tan-1 Â¼ + tan-1 (2/9) = Â½ cos-1 (3/5) = Â½ sin-1 (4/5)

Hence, proved.

(iii) Given tan-1 (2/3) = Â½ tan-1 (12/5)

Hence, proved.

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

Hence, proved.

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

(vi) Given 2 sin-1 (3/5) â€“ tan-1 (17/31) = Ï€/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) = Ï€/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) = Ï€/4

Hence, proved.

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

Solution:

Given sin-1 (2a/1 + a2) â€“ cos-1(1 â€“ b2/1 + b2) = tan-1(2x/1 â€“ x2)

Hence, proved.

4. Prove that:

(i) tan-1{(1 â€“ x2)/ 2x)} + cot-1{(1 – x2)/ 2x)} = Ï€/2

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

Solution:

(i) Given tan-1{(1 â€“ x2)/ 2x)} + cot-1{(1 – x2)/ 2x)} = Ï€/2

Hence, proved.

(ii) Given sin {tan-1 (1 â€“ 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 â€“ 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

## Frequently Asked Questions on RD Sharma Solutions for Class 12 Maths Chapter 4

### Why should I access RD Sharma Solutions for Class 12 Maths Chapter 4?

The reasons why students should access RD Sharma Solutions for Class 12 Maths Chapter 4 are as follows:

1. The CBSE board suggests students practice RD Sharma textbooks on a regular basis as it is one of the best study sources from an exam point of view.
2. RD Sharma Solutions is a key source to procure high marks in exams as the answers for each question are explained in a step-wise manner.Â
3. The subject matter experts and faculty at BYJUâ€™S explain the concepts in a concise manner to help students to speed up their problem-solving skills which are vital from an exam perspective.

### Are RD Sharma Solutions for Class 12 Maths Chapter 4 is enough for effective exam preparation?

Yes, RD Sharma Solutions for Class 12 Maths Chapter 4 prepared by expert teachers boost fundamental concepts among students and strengthen their foundation in basic topics. The solutions are explained in a simple language so that students clear their doubts fastly which arise while solving textbook problems. The main purpose of these solutions is to improve the skills which are essential to score high marks in exams.

### How can we score high marks in RD Sharma Solutions for Class 12 Maths Chapter 4?

RD Sharma Solutions for Class 12 Maths Chapter 4 offers in-depth knowledge of concepts as the answers are well structured by BYJUâ€™S experts based on studentâ€™s intelligence quotient. Every minute detail is explained in a concise manner which helps students grasp the various methods of solving difficult problems more accurately. This also helps students in doing their assignments given to them on time effortlessly.

### Why should I download RD Sharma Solutions for Class 12 Maths Chapter 4?

RD Sharma Solutions offers precise answers for each question of textbook curated by a set of expert faculty at BYJUâ€™S having vast experience in the respective subject. Students are advised to download the solutions in PDF format in order to strengthen their basic concepts in Mathematics. Each solution is explained in a comprehensive manner to make the learning process easy and interesting. Following these solutions help students to obtain an idea of solving complex problems effectively in a short duration of time.

### Does RD Sharma Solutions for Class 12 Maths Chapter 4 boost problem-solving skills among CBSE students?

Yes, RD Sharma Solutions for Class 12 Maths Chapter 4 are created by the highly experienced faculty at BYJUâ€™S based on the latest CBSE syllabus. Students who aim to score more marks are recommended to follow RD Sharma Solutions to obtain an idea of important concepts which are essential from an exam point of view. Using these solutions students can cross-check their answers to know their level of preparation. Regular practice of RD Sharma Solutions helps students to boost problem-solving and time management skills among CBSE students.