RBSE Maths Class 11 Chapter 3: Important Questions and Solutions

RBSE Maths Chapter 3 – Trigonometric functions Class 11 Important questions and solutions are available here. The important questions and solutions of Chapter 3, available at BYJU’S, contain step by step explanations. All these important questions are based on the new pattern prescribed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 11 solutions.

Chapter 3 of the RBSE Class 11 Maths will help the students to solve problems related to angles, signs of trigonometric functions, domain and range of trigonometric functions, a graph of trigonometric functions, trigonometric functions of sum and difference of two angles, trigonometric equations.

RBSE Maths Chapter 3: Miscellaneous Exercise Questions and Solutions

Question 1: A right angle is:

(B) equal to 90 degree

(C) equal to 18°

Solution:

(B) equal to 90 degree

Question 2: Which trigonometric function is positive in the third quadrant?

(A) sin θ

(B) tan θ

(C) cos θ

(D) sec θ

Solution:

(B) tan θ

Question 3: cosec (- θ) is equal to:

(A) sin θ

(B) tan θ

(C) cos θ

(D) -cosec θ

Solution:

(D) -cosec θ

Question 4: tan (90° – θ) is equal to:

(A) -tan θ

(B) cot θ

(C) tan θ

(D) -cot θ

Solution:

(B) cot θ

Question 5: If cos θ = [1 / 2] then value of θ will be:

(A) 2π / 3

(B) π / 3

(C) -2π / 3

(D) 3π / 4

Solution:

cos θ = [1 / 2]

θ = cos-1 (½)

θ = π / 3 or 60o

(B) π / 3

Question 6: If n is an even integer, then the value of sin (2nπ ± θ) will be:

(A) ± cos θ

(B) ± tan θ

(C) ± sin θ

(D) ± cot θ

Solution:

(C) ± sin θ

Question 7: The value of cot 15° will be:

(A) 2 + √3

(B) – 2 + √3

(C) 2 – √3

(D) – 2 – √3

Solution:

Question 8: The value of cos 15° will be:

(A) [√3 + 1] / [2√2]

(B) [√3 – 1] / [2]

(C) [√3 – 1] / [2√2]

(D) [√3 + 1] / [2]

Solution:

(A) [√3 + 1] / [2√2]

cos (a – b) = cos a * cos b + sin a * sin b

cos 15 = cos (45 –30)

=> cos 15 = cos 45 * cos 30 + sin 45 * sin 30

=> cos 15 = (1 / √2) * (√3 / 2) + (1 / √2) * (1 / 2)

=> cos 15 = (√3 + 1) / 2√2

Question 9: The value of 2 sin [5π / 12] * cos [π / 12] will be

(A) 1

(B) √3 / 2

(C) (√3 / 2) – 1

(D) (√3 / 2) + 1

Solution:

Question 10: The value of cos [5π / 12] * sin [π / 12] will be

(A) 1 / 2√2

(B) 0

(C) -2√2

(D) [2 – √3] / 4

Solution:

Question 11: If sin A = 3 / 5, then sin 2A will be:

(A) 3 / 5

(B) 5 / 25

(C) 24 / 25

(D) 4 / 5

Solution:

sin 2A = 2 sin A cos A

= 2 sin A √1 – sin2 A

= 2 * [3 / 5] * √1 – (3 / 5)2

= [6 / 5] * [√(25 – 9) / 25]

= [6 / 5] * [√16 / 25]

= [6 / 5] * [4 / 5]

= 24 / 25

Question 12: If sin A = 3 / 4, then the value of sin 3A will be:

(A) 9 / 16

(B) -9 / 16

(C) 9 / 32

(D) 7 / 16

Solution:

sin 3a = 3 sin A – 4 sin3 A

= 3 * [3 / 4] – 4 * [3 / 4]3

= [9 / 4] – 4 * [27 / 64]

= [9 / 4] * [108 / 64]

= 9 / 16

Question 13: If tan A = 1 / 5, then the value of tan 3A will be:

(A) 47 / 25

(B) 37 / 55

(C) 37 / 11

(D) 47 / 55

Solution:

tan 3A = [3 tan A – tan3 A] / [1 – 3 tan2 A]

= [3 * (1 / 5) – (1 / 5)3] / [1 – 3 * (1 / 5)2]

= [(3 / 5) – (1 / 125)] / [1 – (3 / 25)]

= (74 / 125) / (22 / 25)

= 37 / 55

Question 14: If A + B = π / 4, then the value of (1 + tan A) * (1 + tan B) will be:

(A) 3

(B) 2

(C) 4

(D) 1

Solution:

(1 + tan A) (1 + tan B) = 1 + tan A + tan B + tan A tan B ….. (i)

A + B = π / 4

tan (A + B) = tan [π / 4] = 1

[tan A + tan B] / [1 – tan A tan B] = 1

tan A + tan B = 1 – tan A tan B

1 + tan A + tan B + tan A tan B = 1 + 1 = 2 ….. (ii)

From (i) and (ii),

(1 + tan A) (1 + tan B) = 2

Question 15: The general value of θ in the equation sec2 θ = 2 will be:

(A) 2nπ ± [π / 4]

(B) nπ ± [π / 2]

(C) nπ ± [-1]n [π / 4]

(D) nπ ± [π / 4]

Solution:

sec2 θ = 2

cos2 θ = ½

cos θ = 1 / √2

So, the principal value of θ = 1 / √2.

Choice A is the right answer.

Question 16: Prove the following.

[i] cos θ + sin (27θ° + θ) – sin (27θ° – θ)+ cos (18θ° + θ) = θ

[ii] sec (3π / 2 – θ) sec (θ – 5π / 2) + tan (5π / 2 + θ) tan (θ – 3π / 2) = -1

Solution:

[i] sin (27θ° + θ) → IV quadrant, negative = – cosθ … (i)

sin (27θ° – θ) → III quadrant, negative = – cos θ …. (ii)

cos (18θ° + θ) → III quadrant, negative = – cos θ … (iii)

LHS = cos θ + sin (27θ° + θ) – sin (27θ° – θ) + cos (18θ° + θ)

= cos θ – cos θ + cos θ – cos θ

From (i), (ii) and (iii),

= θ

= RHS

[ii] LHS = sec (3π / 2 – θ) sec (θ – 5π / 2) + tan (5π / 2 + θ) tan (θ – 3π / 2)

= sec (270° – θ) sec (θ – 450°)+ tan (450° + θ) tan (θ – 270°)

= – cosec θ sec (θ – 450°)+ tan (360° + 90° + θ) tan (θ – 270°)

= – cosec θ sec (450° – θ) – tan (90° + θ) tan (270° – θ)

= – cosec θ sec (360° + 90° – θ) – (– cot θ) cot θ

= – cosec θ sec (90° – θ) + cot2 θ

= – cosec θ cosec θ + cot2 θ

= – cosec2 θ + cot2 θ

= – 1 (∵ 1 + cot2 θ = cosec2 θ , cot2 θ – cosec2 θ = – 1)

= RHS

Question 17: Find the value of sin [nπ + (-1)n (π / 4)] , where n is an integer.

Solution:

sin [nπ + (-1)n (π / 4)]

When n = 0, then

sin [0 * π + (-1)0 (π / 4)] = sin (π / 4) = 1 / √2

When n = 1, then

sin [1 * π + (-1)1 (π / 4)] = sin (π – π / 4) = sin (π / 4) = 1 / √2

When n = 2, then

sin [2 * π + (-1)2 (π / 4)] = sin (2π + π / 4) = sin (π / 4) = 1 / √2

When n = 3, then

sin [3 * π + (-1)3 (π / 4)] = sin (3π – π / 4) = sin (π / 4) = 1 / √2

Hence, for n = 0, 1, 2, 3, ……, sin [nπ + (-1)n (π / 4)] = 1 / √2, where n is an integer.

Question 18: If sin A + sin 5 = a and cos A + cos 5 = b, then prove that:

(i) sin (A + B) = [2ab] / [a2 + b2]

(ii) cos (A + B) = [b2 – a2] / [a2 + b2]

Solution:

Question 19: If A + B + C = 180°, then prove that:

(i) cos 2A + cos 2B – cos 2C = 1 – 4 sin A sin B cos C

(ii) sin A + sin B + sin C = 4 sin [A / 2] cos [B / 2] sin [C / 2]

Solution:

(i) L.H.S.

= cos 2A + cos 2B – cos 2C

= (cos 2A + cos 2B) – cos 2C

= 2 cos (A + B) cos (A – B) – cos 2C

= 2 cos (180° – C) cos (A – B) – (2 cos2 C – 1)

= – 2 cos C cos (A – B) – 2 cos2C + 1

= 1 – 2 cos C cos (A – B) – 2 cos2C

= 1 – 2 cos C [cos (A – B) + cos C]

= 1 – 2 cos C ([2 cos (A – B + C)] / 2) * (cos (A – B – C) / 2)

= 1 – 4 cos C [sin B cos [-(90o – A)]

= 1 – 4 cos C sin B cos (90o – A)

= 1 – 4 cos C sin B sin A

= 1 – 4 sin A sin B cos C

= RHS

(ii) (sin A + sin B) + sin C = 2 sin((A + B) / 2) cos ((A – B) / 2) + sin C….(*).

A + B + C = π

A + B = π – C

((A + B) / 2) = (π – C) / 2

((A + B) / 2) = [π / 2] – [C / 2]

sin ((A + B) / 2) = sin (π / 2 – C / 2)

= cos (C / 2) ….(*1)

Also, sinC = 2 sin (C / 2) cos (C / 2) ………(*2).

Utilising (*1) and (*2) in (star), we get,

sinA + sinB + sinC

= 2 cos (C / 2) cos ((A – B ) / 2) + 2 sin (C / 2) cos (C / 2)

= 2 cos (C / 2) {cos ((A – B) / 2) + sin (C / 2)}

= 2 cos (C / 2) {cos ((A – B) / 2) + sin((π – (A + B)) / 2) [as, A + B + C = π],

= 2 cos (C / 2) {cos ((A – B) / 2) + sin (π / 2 – (A + B) / 2)}

= 2 cos (C / 2) {cos((A – B) / 2) + cos ((A + B) / 2)}

= 2 cos (C / 2) {2 cos(((A – B) / 2 + (A + B) / 2) / 2) cos(((A – B) / 2 – (A + B) / 2) / 2)

= 2 cos (C / 2) {2 cos (A / 2) cos (-B / 2)}

= 4cos (A / 2) cos (B / 2) cos (C / 2)….[because, cos(-x)=cosx].

Question 20: If A + B + C = 2π, then prove that cos2B + cos2C – sin2A = 2 cos A cos B cos C.

Solution:

We know that:

cos2 x = [1 + cos 2x] / [2] ; sin2x = [1 + cos 2x] / [2]

LHS = cos2B + cos2C – sin2A

= [1 + cos2 B] / [2] + [1 + cos2 C] / [2] – [1 – cos2 A] / [2]

= [1 / 2] * [2 cos2 B] + 2 cos (A + B) cos (A – C)

= = [cos2B + cos (2π + B) cos (A – C)]

= cos2B + cos B cos (A – C)

= cos B (cos B + cos (A – C)

= cos B (cos (2π – (A – C) + cos (A – C) – cos B [cos (A – B) + cos (A + C)]

= cos B [(cos (A + B) + cos (A – C)]

= cos B [2 cos A cos C]

= 2 cos A cos B cos C

Question 21: Find the following equation 2 tan θ – cot θ + 1 = 0.

Solution: