 # RBSE Maths Class 10 Chapter 6: Trigonometric Ratios Important Questions and Solutions

RBSE Class 10 Maths Chapter 6 – Trigonometric Ratios Important questions and solutions are given here. All these questions have detailed solutions with accurate answers. The RBSE Class 10 important questions and solutions available at BYJU’S contain all the topics of Class 10 and are given as per the new guidelines of the board.

Chapter 6 of RBSE Class 10 has only one exercise, which covers the questions about table values of sine, cosine, tangent, secant, cosecant, and cotangent for acute angles. Enough questions are provided here so that students will be able to solve any type of question under this category after practising all these questions. Besides, there exist some questions on proving the identities, which involve ratios of trigonometry.

### RBSE Maths Chapter 6: Exercise 6.1 Textbook Important Questions and Solutions

Question 1: Find the value of 2 sin 45° cos 45°.

Solution:

2 sin 45° cos 45°

= 2 × (1/√2) × (1/√2)

= 2 × (1/2)

= 1

Question 2: Find the value of cos 45° cos 60° – sin 45° sin 60°.

Solution:

cos 45° cos 60° – sin 45° sin 60°

= (1/√2) × (1/2) – (1/√2) × (√3/2)

= (1/2√2) – (√3/2√2)

= (1 – √3)/2√2

Question 3: Evaluate: sin230° + 2 cos245° + 3 tan260°

Solution:

sin230° + 2 cos245° + 3 tan260°

= (1/2)2 + 2(1/√2)2 + 3(√3)2

= (1/4) + 2(1/2) + 3(3)

= (1/4) + 1 + 9

= 10 + 1/4

= (40 + 1)/4

= 41/4

Question 4: Find the value of 3 sin 60° – 4 sin360°

Solution:

3 sin 60° – 4 sin360°

= 3(√3/2) – 4(√3/2)3

= (3√3)/2 – (4 × 3√3)/8

= (3√3)/2 – (3√3)/2

= 0

Question 5: Evaluate: (5 cos260° + 4 sin230° + tan245°)/ (sin230° + cos245°)

Solution:

(5 cos260° + 4 sin230° + tan245°)/ (sin230° + cos245°)

= [5(1/2)2 + 4(1/2)2 + (1)2] / [(1/2)2 + (1/√2)2]

= [ 5(1/4) + 4(1/4) + 1] / [(1/4) + (1/2)]

= [(5/4) + 1 + 1]/ [(1 + 2)/4]

= [(5 + 4 + 4)/4]/(3/4)

= (13/4) × (4/3)

= 13/3

Question 6: Find the value of 4 cot245° – sec260° + sin260° + cos290°.

Solution:

4 cot245° – sec260° + sin260° + cos290°

= 4(1)2 – (2)2 + (√3/2)2 + (0)2

= 4 – 4 + (3/4) + 0

= 3/4

Question 7: Simplify: (4/cot230°) + (1/sin230°) – cos245°

Solution:

(4/cot230°) + (1/sin230°) – cos245°

= 4/(√3)2 + 1/(1/2)2 – (1/√2)2

= (4/3) + 4 – (1/2)

= (8 + 24 – 3)/6

= 29/6

Question 8: Evaluate: (tan260° + 4sin245° + sin290°) / (3 sec230° + cosec260° – cot230°)

Solution:

(tan260° + 4sin245° + sin290°) / (3 sec230° + cosec260° – cot230°)

= [(√3)2 + 4(1/√2)2 + (1)2] / [3(2/√3)2 + (2/√3)2 – (√3)2]

= [3 + 4(1/2) + 1] / [3(4/3) + (4/3) – 3]

= (4 + 2)/ (1 + 4/3)

= 6/ [(3 + 4)/3]

= (6 × 3)/7

= 18/7

Question 9: Find the value of (sin 30° – sin 90° + 2 cos 0°)/(tan 30° tan 60°)

Solution:

(sin 30° – sin 90° + 2 cos 0°)/(tan 30° tan 60°)

= [(1/2) – 1 + 2(1)] / [(1/√3) × (√3)]

= [(1/2) + 1]/1

= (1 + 2)/2

= 3/2

Question 10: Simplify: 2 tan 30°/ (1 – tan230°)

Solution:

2 tan 30°/ (1 – tan230°)

= [2 (1/√3)] / [1 – (1/√3)2]

= (2/√3) / (1 – 1/3)

= (2/√3)/ [(3 – 1)/3]

= (2/√3) × (3/2)

= 3/√3

= √3

Question 11: Find the value of x, if sin 2x = sin 60° cos 30° – cos 60° sin 30°

Solution:

sin 2x = sin 60° cos 30° – cos 60° sin 30°

= (√3/2) × (√3/2) – (1/2) × (1/2)

= (3/4) – (1/4)

= (3 – 1)/4

= 2/4

= 1/2

sin 2x = sin 30°

2x = 30°

⇒ x = 30°/2

⇒ x = 15°

Question 12: Prove that 4 cot245° – sec260° – sin230° = -1/4

Solution:

LHS = 4 cot245° – sec260° – sin230°

= 4(1)2 – (2)2 – (1/2)2

= 4 – 4 – 1/4

= -1/4

= RHS

Therefore, 4 cot245° – sec260° – sin230° = -1/4

Hence proved.

Question 13: Prove that: 4 sin 30° sin260° + 3 cos 60° tan 45° = 2 sec245° – cosec290°

Solution:

LHS = 4 sin 30° sin260° + 3 cos 60° tan 45°

= 4 × (1/2) × (√3/2)2 + 3 × (1/2) × (1)

= 2 × (3/4) + (3/2)

= (3/2) + (3/2)

= 2 × (3/2)

= 3

RHS = 2 sec245° – cosec290°

= 2(√2)2 – (1)2

= 2(2) – 1

= 4 – 1

= 3

Therefore, LHS = RHS

i.e. 4 sin 30° sin260° + 3 cos 60° tan 45° = 2 sec245° – cosec290°

Hence proved.

Question 14: Prove that: (sin 60° + sin 30°)/(sin 60° – sin 30°) = (tan 60° + tan 45°)/(tan 60° – tan 45°)

Solution:

LHS = (sin 60° + sin 30°)/(sin 60° – sin 30°)

= [(√3/2) + (1/2)] / [(√3/2) – (1/2)]

= [(√3 + 1)/2] / [(√3 – 1)/2]

= (√3 + 1)/(√3 – 1)

RHS = (tan 60° + tan 45°)/(tan 60° – tan 45°)

= (√3 + 1)/(√3 – 1)

LHS = RHS

Therefore, (sin 60° + sin 30°)/(sin 60° – sin 30°) = (tan 60° + tan 45°)/(tan 60° – tan 45°)

Hence proved.

Question 15: Prove that: 2(cos245° + tan260°) – 6(sin245° – tan230°) = 6

Solution:

LHS = 2(cos245° + tan260°) – 6(sin245° – tan230°)

= 2[(1/√2)2 + (√3)2] – 6[(1/√2)2 – (1/√3)2]

= 2[(1/2) + 3] – 6[(1/2) – (1/3)]

= 2[(1 + 6)/2] – 6[(3 – 2)/6]

= 7 – 1

= 6

= RHS

Therefore, 2(cos245° + tan260°) – 6(sin245° – tan230°) = 6

Hence proved.

Question 16: Prove that: (1 – sin 45° + sin 30°)(1 + cos 45° + cos 60°) = 7/4

Solution:

LHS = (1 – sin 45° + sin 30°)(1 + cos 45° + cos 60°)

= [1 – (1/√2) + (1/2)] [1 +(1/√2) + (1/2)

= [(3/2) – (1/√2)] [(3/2) + (1/√2)]

= (3/2)2 – (1/√2)2

= (9/4) – (1/2)

= (9 – 2)/4

= 7/4

= RHS

Therefore, (1 – sin 45° + sin 30°)(1 + cos 45° + cos 60°) = 7/4

Hence proved.

Question 17: Prove that: cos20° – 2 cot230° + 3 cosec290° = 2(sec245° – tan260°)

Solution:

LHS = cos20° – 2 cot230° + 3 cosec290°

= (1)2 – 2(√3)2 + 3(1)2

= 1 – 2(3) + 3

= 4 – 6

= -2

RHS = 2(sec245° – tan260°)

= 2[(√2)2 – (√3)2]

= 2(2 – 3)

= 2(-1)

= -2

Therefore, cos20° – 2 cot230° + 3 cosec290° = 2(sec245° – tan260°)

Hence proved.

Question 18: If x = 30°, then prove that: tan 2x = 2 tan x/(1 – tan2x)

Solution:

Given,

x = 30°

LHS = tan 2x

= tan 2(30°)

= tan 60°

= √3

RHS = 2 tan x/(1 – tan2x)2

= 2 tan 30°/(1 – tan230°)

= 2(1/√3)/[1 – (1/√3)]

= (2/√3)/ (1 – 1/3)

= (2/√3)/[(3 – 1)/3]

= (2/√3) × (3/2)

= 3/√3

= √3

Therefore, tan 2x = 2 tan x/(1 – tan2x)

Hence proved.

Question 19: If x = 30°, then prove that: cos 3x = 4 cos3x – 3 cos x

Solution:

Given,

x = 30°

LHS = cos 3x

= cos 3(30°)

= cos 90°

= 0

RHS = 4 cos3x – 3 cos x

= 4 cos330° – 3 cos 30°

= 4(√3/2)3 – 3(√3/2)

= 4(3√3/8) – (3√3/2)

= (3√3/2) – (3√3/2)

= 0

Therefore, cos 3x = 4 cos3x – 3 cos x

Hence proved.

Question 20: If A = 60° and B = 30°, then prove that: cot(A – B) = (cot A cot B + 1)/(cot B – cot A)

Solution:

Given,

A = 60° and B = 30°

LHS = cot(A – B)

= cot(60° – 30°)

= cot 30°

= √3

RHS = (cot A cot B + 1)/(cot B – cot A)

= (cot 60° cot 30° + 1)/(cot 30° – cot 60°)

= [(1/√3) × (√3) + 1] / [√3 – (1/√3)]

= (1 + 1)/ [(3 – 1)/√3]

= 2 × (√3/2)

= √3

Therefore, cot(A – B) = (cot A cot B + 1)/(cot B – cot A)

Hence proved.

### RBSE Maths Chapter 6: Additional Important Questions and Solutions

Question 1: The value of tan260° is:

(a) 3

(b) 1/3

(c) 1

(d) ∞

Solution:

tan260° = (√3)2 = 3

Question 2: The value of 2 sin260° cos 60° will be:

(a) 4/3

(b) 5/2

(c) 3/4

(d) 1/3

Solution:

2 sin260° cos 60°

= 2 × (√3/2)2 × (1/2)

= 2 × (3/4) × (1/2)

= 3/4

Question 3: If cosec θ = 2/√3, then value of θ is:

(a) π/4

(b) π/3

(c) π/2

(d) π/6

Solution:

cosec θ = 2/√3

cosec θ = cosec 60°

= cosec π/3

Therefore, θ = π/3

Question 4: If θ = 45°, then value of (1 – cos 2θ)/sin 2θ is:

(a) 0

(b) 1

(c) 2

(d) ∞

Solution:

Given,

θ = 45°

(1 – cos 2θ)/sin 2θ = [1 – cos 2(45°)]/ sin 2(45°)

= (1 – cos 90°)/sin 90°

= (1 – 0)/1

= 1/1

= 1

Question 5: Prove that: cos 60° = 2 cos230° – 1

Solution:

LHS = cos 60°

= 1/2

RHS = 2 cos230° – 1

= 2(√3/2)2 – 1

= 2(3/4) – 1

= (3/2) – 1

= (3 – 2)/2

= ½

Therefore, cos 60° = 2 cos230° – 1

Hence proved.

Question 6: Prove that: sin 60° = 2 tan 30°/(1 + tan230°)

Solution:

LHS = sin 60° = √3/2

RHS = 2 tan 30°/(1 + tan230°)

= [2 (1/√3)] / [1 + (1/√3)2]

= (2/√3)/ [1 + (1/3)]

= (2/√3)/[(3 + 1)/3]

= (2/√3) × (3/4)

= √3/2

Therefore, sin 60° = 2 tan 30°/(1 + tan230°)

Hence proved.

Question 7: Prove that: cos 60° = (1 – tan230°)/(1 + tan230°)

Solution:

LHS = cos 60° = 1/2

RHS = (1 – tan230°)/(1 + tan230°)

= [1 – (1/√3)2] / [1 + (1/√3)2]

= [1 – (1/3)] / [1 + (1/3)]

= [(3 – 1)/3] / [(3 + 1)/3]

= (2/3) × (3/4)

= ½

Therefore, cos 60° = (1 – tan230°)/(1 + tan230°)

Hence proved.

Question 8: Prove that: (sin 45° + cos 45°)2 = 2

Solution:

LHS = (sin 45° + cos 45°)2

= [(1/√2) + (1/√2)]2

= [(1 + 1)/√2]2

= (2/√2)2

= 4/2

= 2

Therefore, (sin 45° + cos 45°)2 = 2

Hence proved.

Question 9: Prove that: 4 tan 30° sin 45° sin 60° sin 90° = √2

Solution:

LHS = 4 tan 30° sin 45° sin 60° sin 90°

= 4 × (1/√3) × (1/√2) × (√3/2) × (1)

= 2/√2

= √2

= RHS

Therefore, 4 tan 30° sin 45° sin 60° sin 90° = √2

Hence proved.

Question 10: Find the value of sin260° cot260°.

Solution:

sin260° cot260°

= (√3/2)2 × (1/√3)2

= (3/4) × (1/3)

= 1/4

Question 11: If cot θ = 1/√3, then prove that: (1 – cos2θ)/(2 – sin2θ) = 3/5

Solution:

Given,

cot θ = 1/√3

cot θ = cot 60°

⇒ θ = 60°

LHS = (1 – cos2θ)/(2 – sin2θ)

= (1 – cos260°)/(2 – sin260°)

= [1 – (1/2)2] / [2 – (√3/2)2]

= [1 – (1/4)] / [2 – (3/4)]

= [(4 – 1)/4] / [(8 – 3)/4]

= (3/4) × (4/5)

= 3/5

= RHS

Therefore, (1 – cos2θ)/(2 – sin2θ) = 3/5

Hence proved.

Question 12: Prove that: 3(tan230° + cot230°) – 8(sin245° + cos230°) = 0

Solution:

LHS = 3(tan230° + cot230°) – 8(sin245° + cos230°)

= 3[(1/√3)2 + (√3)2] – 8[(1/√2)2 + (√3/2)2]

= 3[(1/3) + 3] – 8[(1/2) + (3/4)]

= 3[(1 + 9)/3] – 8[(2 + 3)/4]

= 10 – 2(5)

= 10 – 10

= 0

= RHS

Therefore, 3(tan230° + cot230°) – 8(sin245° + cos230°) = 0

Hence proved.

Question 13: Prove that: 4(sin430° + cos 60°) – 3(cos245° – sin290°) = 15/4

Solution:

LHS = 4(sin430° + cos 60°) – 3(cos245° – sin290°)

= 4[(1/2)4 + (1/2)] – 3[(1/√2)2 – (1)2]

= 4[(1/16) + (1/2)] – 3[(1/2) – 1]

= 4[(1 + 8)/16] – 3[(1 – 2)/2]

= (9/4) + (3/2)

= (9 + 6)/4

= 15/4

= RHS

Therefore, 4(sin430° + cos 60°) – 3(cos245° – sin290°) = 15/4

Hence proved.

Question 14: Prove that: (cos 30° + sin 60°)/ (1 + cos 60° + sin 30°) = √3/2

Solution:

LHS = (cos 30° + sin 60°)/ (1 + cos 60° + sin 30°)

= [(√3/2) + (√3/2)] / [1 + (1/2) + (1/2)]

= [(2√3)/2] / [1 + 1]

= √3/2

= RHS

Therefore, (cos 30° + sin 60°)/ (1 + cos 60° + sin 30°) = √3/2

Hence proved.

Question 15: Find the value of tan260° + 3 cos230°.

Solution:

tan260° + 3 cos230° = (√3)2 + 3(√3/2)2

= 3 + 3(3/4)

= 3 + 9/4

= (12 + 9)/4

= 21/4  by India's top teacher with
live doubt solving