 # RBSE Maths Class 12 Chapter 2: Inverse Circular Functions Important Questions and Solutions

RBSE Class 12 Maths Chapter 2 – Inverse Circular Functions Important questions and solutions are available here. All these important questions and solutions are given by our expert faculty, based on the new examination pattern. The RBSE Class 12 important questions and solutions given at BYJU’S, will help the students in clearing doubts and understanding all the concepts in detail.

Chapter 2 of RBSE Class 12 has only one exercise which covers the questions on the principal solution of trigonometric functions using the concepts of principal value, domain and the relation between the functions of circular inverse trigonometry. Adding to these concepts, different types of properties and formulae are given in this chapter which will help in solving various problems in the exam.

### RBSE Maths Chapter 2: Exercise 2.1 Textbook Important Questions and Solutions

Question 1: Find the principal value of the following angles.

(i) sin-1(1)

(ii) sec-1(-√2)

(iii) tan-1(1/√3)

Solution:

(i) We know that, principal value of sin-1(1) will be in the range of [-π/2, π/2]

Let sin-1(1) = x

⇒ sin x = 1

⇒ sin x = sin π/2

⇒ x = π/2

Therefore, the principal value of sin-1(1) is π/2.

(ii) We know that, the principal value of sec-1(-√2) will be in the range of [0, π] – {π/2}

Let sec-1(-√2) = x

sec x = -√2

sec x = -sec π/4

sec x = sec(π – π/4)

sec x = sec(3π/4)

x = 3π/4

Therefore, the principal value of sec-1(-√2) is 3π/4.

(i) We know that the principal value of tan-1(1/√3) will be in the range of [-π/2, π/2].

Let tan-1(1/√3) = x

tan x = 1/√3

tan x = tan π/6

x = π/6

Therefore, the principal value of tan-1(1/√3) is π/6.

Question 2: Prove that: 2 tan-1(1/2) – tan-1(1/7) = π/4

Solution:

We know that,

2 tan-1(x) = tan-1[2x/(1 – x2)]

LHS = 2 tan-1(1/2) – tan-1(1/7)

= tan-1[2×(1/2)/ 1 – (1/2)2] – tan-1(1/7)

= tan-1[1/(1 – 1/4)] – tan-1(1/7)

= tan-1(4/3) – tan-1(1/7)

Using the formula, tan-1(x) – tan-1(y) = tan-1[(x -y)/(1 + xy)]

= tan-1[(28 – 3)/(21 + 4)]

= tan-1(25/25)

= tan-1(1)

= tan-1(tan π/4)

= π/4

= RHS

Hence, proved that 2 tan-1(1/2) – tan-1(1/7) = π/4.

Question 3: Prove that: sec2(tan-12) + cosec2(cot-13) = 15

Solution:

Let tan-12 = θ

⇒ tan θ = 2

We know that,

sec2θ = 1 + tan2θ

= 1 + (2)2

= 1 + 4

= 5

∴ sec2(tan-12) = 5 ….(i)

Let cot-13 = Φ

⇒ cot Φ = 3

∴ cosec2Φ = 1 + cot2Φ

= 1 + (3)2

= 1 + 9

= 10

∴ cosec2(cot-13) = 10 ….(ii)

sec2(tan-12) + cosec2(cot-13) = 5 + 10

sec2(tan-12) + cosec2(cot-13) = 15

Hence proved.

Question 4: If cos-1x + cos-1y + cos-1z = π, then prove that x2 + y2 + z2 + 2xyz = 1.

Solution:

Given,

cos-1x + cos-1y + cos-1z = π

cos-1x + cos-1y = π – cos-1z

⇒ cos-1[xy – √(1 – x2) √(1 – y2)] = cos-1(-z)

⇒ xy – √(1 – x2) √(1 – y2) = (-z)

⇒ xy + z = √(1 – x2) √(1 – y2)

Squaring on both sides,

⇒ (xy + z)2 = [√(1 – x2) √(1 – y2)]2

⇒ x2y2 + z2 + 2xyz = (1 – x2)(1 – y2)

⇒ x2y2 + z2 + 2xyz = 1 – x2 – y2 + x2y2

⇒ z2 + 2xyz = 1 – x2 – y2

⇒ x2 + y2 + z2 + 2xyz = 1

Hence proved.

Question 5: If tan-1x + tan-1y + tan-1z = π/2, then prove that xy + yz + zx = 1.

Solution:

Given,

tan-1x + tan-1y + tan-1z = π/2

tan-1[(x + y + z – xyz)/ (1 – xy – yz – xz)] = π/2

(x + y + z – xyz)/ (1 – xy – yz – xz) = tan π/2

(x + y + z – xyz)/ (1 – xy – yz – xz) = ∞

or

(x + y + z – xyz)/ (1 – xy – yz – xz) = 1/0

⇒ 1 – xy – yz – xz = 0

⇒ xy + yz + xz = 1

Hence proved.

Question 6: Prove that: tan-1x + cot-1(x + 1) = tan-1(x2 + x + 1)

Solution:

LHS = tan-1x + cot-1(x + 1)

= tan-1x + tan-1[1/(x + 1)]

= tan-1{[x + (1/x + 1)]/[1 – x(1/x + 1)]}

= tan-1[(x2 + x + 1)/(x + a – x)]

= tan-1(x2 + x + 1)

= RHS

Hence, proved that: tan-1x + cot-1(x + 1) = tan-1(x2 + x + 1)

Question 7: If tan-1x, tan-1y, tan-1z are in AP, then prove that: y2(x + z) + 2y(1 – xz) – x – z = 0.

Solution:

Given,

tan-1x, tan-1y, tan-1z are in AP.

tan-1y = (tan-1x + tan-1z)/2

⇒ tan-1x + tan-1z = 2 tan-1y

⇒ tan-1[(z + x)/(1 – zx)] = 2 tan-1y

⇒ tan-1[(z + x)/(1 – zx)] = tan-1[2y/(1 – y2)]

⇒ (z + x)/(1 – zx) = 2y/(1 – y2)

⇒ (z + x)(1 – y2) = 2y(1 – zx)

⇒ z + x – y2(x + z) = 2y(1 – zx)

⇒ y2(x + z) + 2y(1 – zx) – x – z = 0

Hence proved.

Question 8: If the roots of x3 + px2 + qx + p = 0 are α, β, γ, then prove that (except one situation) tan-1α + tan-1β + tan-1γ = nπ and also find the situation when it does not happen.

Solution:

Given,

α, β, γ are the roots of the equation x3+ px2 + qx + p = 0.

α + β + γ = -(coefficient of x2)/(coefficient of x3)

= -p/1

= -p

αβ + βγ + γα = (coefficient of x)/(coefficient of x3)

= q/1

= q

αβγ = -(constant term)/(coefficient of x3)

= -p/1

= -p

LHS = tan-1α + tan-1β + tan-1γ

= tan-1[(α + β + γ – αβγ)/ 1 – (αβ + βγ + γα)]

= tan-1[(-p + p)/(1 – q)]

= tan-1(0)

= nπ

= RHS

Hence proved.

Special Situation:

If the sum and product of roots is not equal, then tan-1α + tan-1β + tan-1γ ≠ nπ.

Question 9: Solve: cos-1[(x2 – 1)/(x2 + 1)] + tan-1[2x/(x2 – 1)] = 2π/3

Solution:

Given,

cos-1[(x2 – 1)/(x2 + 1)] + tan-1[2x/(x2 – 1)] = 2π/3

We know that,

cos-1(-x) = π – cos-1(x)

and

tan-1(-x) = -tan-1(x)

π – cos-1[(1 – x2)/(x2 + 1)] – tan-1[2x/(1 – x2)] = 2π/3

⇒ π – 2 tan-1(x) – 2 tan-1(x) = 2π/3

⇒ 4 tan-1(x) = π – (2π/3)

⇒ 4 tan-1(x) = (3π – 2π)/3

⇒ 4 tan-1(x) = π/3

⇒ tan-1(x) = π/12

⇒ x = tan(π/12)

Question 10: Solve: sin-1(1/√5) + cot-1(x) = π/4

Solution:

Given,

sin-1(1/√5) + cot-1(x) = π/4

We know that,

sin-1(1/x) = y = cosec-1(x)

Thus, sin-1(1/√5) = cosec-1(√5)

Let cosec-1(√5) = y

⇒ cosec y = √5

Squaring on both sides,

cosec2y = (√5)2

cosec2y = 5

Using the identity cosec2A – cot2A = 1,

1 + cot2y = 5

cot2y = 5 – 1

cot2y = 4

⇒ cot y = ±2

Let us take the positive value of cot y.

y = cot-1(2)

Thus, y + cot-1(x) = π/4

⇒ cot-1(2) + cot-1(x) = π/4

⇒ cot-1[(2x – 1)/(2 + x)] = π/4

⇒ (2x – 1)/(2 + x) = cot (π/4)

⇒ (2x – 1)/(2 + x) = 1

⇒ 2x – 1 = 2 + x

⇒ 2x – x = 2 + 1

⇒ x = 3

### RBSE Maths Chapter 2: Additional Important Questions and Solutions

Question 1: The principal value of tan-1(-1) is

(a) 45°

(b) 135°

(c) -45°

(d) -60°

Solution:

We know that,

tan-1(-x)= -tan-1x

Thus, tan-1(-1) = -tan-1(1)

Let tan-1(1) = θ

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°

Therefore, tan-1(-1) = -45°

Question 2: The value of cot(tan-1α + cot-1α) is

(a) 1

(b) ∞

(c) 0

(d) None of these

Solution:

We know that,

tan-1x + cot-1x = π/2

cot(tan-1α + cot-1α) = cot π/2 = 0

Question 3: If tan-1(3x) + tan-1(2x) = π/4, then the value of x is

(a) 1/6

(b) 1/3

(c) 1/10

(d) 1/2

Solution:

tan-1(3x) + tan-1(2x) = π/4

tan-1[(3x + 2x)/ (1 – 3x.2x)] = π/4

5x/(1 – 6x2) = tan π/4

5x/(1 – 6x2) = 1

⇒ 1 – 6x2 = 5x

⇒ 6x2 + 5x – 1 = 0

⇒ 6x2 + 6x – x – 1 = 0

⇒ 6x(x + 1) – 1(x + 1) = 0

⇒ (x + 1)(6x – 1) = 0

⇒ x = -1, x = 1/6

Question 4: If tan-1(1) + cos-1(1/√2) = sin-1(x), then the value of x is

(a) -1

(b) 0

(c) 1

(d) -1/2

Solution:

Given,

tan-1(1) + cos-1(1/√2) = sin-1(x)

tan-1(tan π/4) + cos-1(cos π/4) = sin-1x

(π/4) + (π/4) = sin-1x

⇒ sin-1x = (2π/4)

⇒ sin-1x = π/2

⇒ x = sin π/2

⇒ x = 1

Question 5: If sin-1(3/4) + sec-1(4/3) = x, then find the value of x.

Solution:

Given,

sin-1(3/4) + sec-1(4/3) = x

⇒ sin-1(3/4) + cos-1(3/4) = x

We know that, sin-1x + cos-1x = π/2

⇒ π/2 = x

Therefore, x = π/2