RBSE Class 10 Maths Chapter 7 – Trigonometric Identities Important questions and solutions are provided here. All these questions have stepwise solutions and have given the required reasons for critical steps so that students will be able to understand clearly. The RBSE Class 10 important questions and solutions given at BYJU’S cover all the concepts of Class 10, which will help the students to maximize their marks.
Chapter 7 of RBSE Class 10 has two exercises which contain questions involving identities and complementary angles of trigonometry. Several questions are provided here, which cover proving trigonometric expressions with the help of identities and complementary angles. Also, students will be introduced to the concept of mutual relations between trigonometric ratios.
RBSE Maths Chapter 7: Exercise 7.1 Textbook Important Questions and Solutions
Question 1: Express all trigonometric ratios in terms of sec θ, for angle θ.
Solution:
We know that,
sin2θ + cos2θ = 1
sin2θ = 1 – cos2θ
sin θ = (1 – cos2θ)
= √[1 – (1/sec2θ)]
= √[(sec2θ – 1)/sec2θ]
= √(sec2θ – 1) / sec θ
cos θ = 1/sec θ
1 + tan2θ = sec2θ
tan2θ = sec2θ – 1
tan θ = √(sec2θ – 1)
cosec θ = 1/sin θ
= sec θ/√(sec2θ – 1)
cot θ = 1/tan θ = 1/√(sec2θ – 1)
Question 2: Prove that cos2θ + cos2θ cot2θ = cot2θ
Solution:
LHS = cos2θ + cos2θ cot2θ
= cos2θ (1 + cot2θ)
Using the identity 1 + cot2A = cosec2A
= cos2θ × cosec2θ
= cos2θ × (1/sin2θ)
= cot2θ
= RHS
Therefore, cos2θ + cos2θ cot2θ = cot2θ
Hence proved.
Question 3: Prove that sec θ(1 – sin θ)(sec θ + tan θ) = 1 using the identities.
Solution:
LHS = sec θ(1 – sin θ)(sec θ + tan θ)
= (1/cos θ) (1 – sin θ) [(1/cos θ) + (sin θ/cos θ)]
= (1/ cos θ) (1 – sin θ)[(1 + sin θ)/cos θ]
= (1 – sin2θ)/cos2θ
Using the identity sin2A + cos2A = 1,
= cos2θ/cos2θ
= 1
= RHS
Therefore, sec θ(1 – sin θ)(sec θ + tan θ) = 1
Hence proved.
Question 4: Prove that cosec2θ + sec2θ = cosec2θ sec2θ with the help of identities.
Solution:
LHS = cosec2θ + sec2θ
= (1/sin2θ) + (1/cos2θ)
= (cos2θ + sin2θ) / (sin2θ cos2θ)
Using the identity sin2A + cos2A = 1,
= (1/sin2θ) (1/cos2θ)
= cosec2θ sec2θ
= RHS
Therefore, cosec2θ + sec2θ = cosec2θ sec2θ
Hence proved.
Question 5: Prove that: (tan A + tan B)/(cot A + cot B) = tan A tan B
Solution:
LHS = (tan A + tan B)/(cot A + cot B)
= [(sin A/cos A) + (sin B/cos B)] / [(cos A/sin A) + (cos B/sin B)]
= [(sin A cos B + cos A sin B) / cos A cos B] / [(cos A sin B + cos B sin A) / sin A sin B]
= (sin A sin B/cos A cos B) × [(sin A cos B + cos A sin B)/(sin A cos B + cos A sin B)]
= (sin A/cos A) × (sin B/cos B)
= tan A tan B
= RHS
Therefore, (tan A + tan B)/(cot A + cot B) = tan A tan B
Hence proved.
Question 6: Prove that: (sin4θ – cos4θ)/(sin2θ – cos2θ) = 1 with the help of identities.
Solution:
LHS = (sin4θ – cos4θ)/(sin2θ – cos2θ)
= [(sin2θ)2 – (cos2θ)2] / (sin2θ – cos2θ)
= [(sin2θ + cos2θ)(sin2θ – cos2θ)]/ (sin2θ – cos2θ)
= sin2θ + cos2θ
Using the identity sin2A + cos2A = 1,
= 1
= RHS
Therefore, (sin4θ – cos4θ)/(sin2θ – cos2θ) = 1
Hence proved.
Question 7: Prove that cot θ – tan θ = (1 – 2 sin2θ)/ sin θ cos θ.
Solution:
LHS = cot θ – tan θ
= (cos θ/sin θ) – (sin θ/cos θ)
= (cos2θ – sin2θ)/sin θ cos θ
Using the identity, sin2A + cos2A = 1,
= [(1 – sin2θ) – sin2θ]/sin θ cos θ
= (1 – 2 sin2θ)/sin θ cos θ
= RHS
Therefore, cot θ – tan θ = (1 – 2 sin2θ)/ sin θ cos θ
Hence proved.
Question 8: Prove that cos4θ + sin4θ = 1 – 2 cos2θ sin2θ
Solution:
LHS = cos4θ + sin4θ
= (cos2θ)2 + (sin2θ)2 + 2 sin2θ cos2θ – 2 sin2θ cos2θ
= (sin2θ + cos2θ)2 – 2 sin2θ cos2θ
Using the identity sin2A + cos2A = 1,
= 1 – 2sin2θ cos2θ
= RHS
Therefore, cos4θ + sin4θ = 1 – 2 cos2θ sin2θ
Hence proved.
Question 9: Prove that (sec θ – cos θ)(cot θ + tan θ) = tan θ sec θ
Solution:
LHS = (sec θ – cos θ)(cot θ + tan θ)
= [(1/cos θ) – cos θ] [(cos θ/sin θ) + (sin θ/cos θ)]
= [(1 – cos2θ)/cos θ] [(cos2θ + sin2θ)/sin θ cos θ]
Using the identity sin2A + cos2A = 1,
= (sin2θ/cos θ) (1/sin θ cos θ)
= (sin θ/cos θ) (1/cos θ)
= tan θ sec θ
= RHS
Therefore, (sec θ – cos θ)(cot θ + tan θ) = tan θ sec θ
Hence proved.
Question 10: Prove that: (1 – tan2α) / (cot2α – 1) = tan2α
Solution:
LHS = (1 – tan2α) / (cot2α – 1)
= (1 – tan2α)/ [(1/tan2α) – 1]
= (1 – tan2α)/ [(1 – tan2α) / tan2α]
= [(1 – tan2α)/ (1 – tan2α)] × tan2α
= tan2α
= RHS
Hence proved.
Question 11: Prove that sin θ/(1 – cos θ) = (1 + cos θ)/sin θ with help of identities.
Solution:
LHS = sin θ/(1 – cos θ)
By rationalising the denominator,
= [sin θ/(1 – cos θ)] [(1 + cos θ)/(1 + cos θ)]
= [sin θ(1 + cos θ)]/(1 – cos2θ)
Using the identity sin2A + cos2A = 1,
= [sin θ(1 + cos θ)]/sin2θ
= (1 + cos θ)/sin θ
= RHS
Hence proved.
Question 12: Prove that: sin6θ + cos6θ = 1 – 3 sin2θ cos2θ
Solution:
LHS = (sin2θ)3 + (cos2θ)3
= (sin2θ + cos2θ) [(sin2θ)2 + (cos2θ)2 – sin2θ cos2θ]
Using the identity sin2A + cos2A = 1,
= (1) (sin4θ + cos4θ – sin2θ cos2θ)
= (sin2θ)2 + (cos2θ)2 + 2 sin2θ cos2θ – 2 sin2θ cos2θ – sin2θ cos2θ
= (sin2θ + cos2θ)2 – 3 sin2θ cos2θ
= (1)2 – 3 sin2θ cos2θ
= 1 – 3 sin2θ cos2θ
= RHS
Hence proved.
Question 13: Prove that: [tan A/(1 – cot A)] + [cot A/(1 – tan A)] = 1 + tan A + cot A
Solution:
tan A(1 – cot A) = (sin A/cos A)/[1 – (cos A/sin A)]
= (sin A/cos A)/ [(sin A – cos A)/sin A]
= sin2A/ [cos A(sin A – cos A)]
cot A/(1 – tan A) = (cos A/sin A)/ [1 – (sin A/cos A)]
= (cos A/sin A)/ [(cos A – sin A)/cos A]
= cos2A/ [sin A(cos A – sin A)]
Now,
LHS = [tan A/(1 – cot A)] + [cot A/(1 – tan A)]
= sin2A/ [cos A(sin A – cos A)] + cos2A/ [sin A(cos A – sin A)]
= sin2A/ [cos A(sin A – cos A)] – cos2A/ [sin A(sin A – cos A)]
= [1/(sin A – cos A)] × [(sin2A/cos A) – (cos2A/sin A)]
= [1/(sin A – cos A)]× [(sin3A – cos3A)/sin A cos A]
= [(sin A – cos A)(sin2A + cos2A + sin A cos A)]/ (sin A – cosA) sin A cos A
= (sin2A + cos2A + sin A cos A)/sin A cos A
= (sin2A/sin A cos A) + (cos2A/sin A cos A) + (sin A cos A/sin A cos A)
= tan A + cot A + 1
= RHS
Hence proved.
Question 14: Prove that: sin θ(1 + tan θ) + cos θ(1 + cot θ) = cosec θ + sec θ
Solution:
LHS = sin θ(1 + tan θ) + cos θ(1 + cot θ)
= sin θ[1 + (sin θ/cos θ)] + cos θ[1 + (cos θ/sin θ)]
= sin θ[(cos θ + sin θ)/cos θ] + cos θ[(sin θ + cos θ)/sin θ]
= (sin θ + cos θ) × [(sin θ/cos θ) + (cos θ/sin θ)]
= (sin θ + cos θ)[(sin2θ + cos2θ)/sin θ cos θ]
Using the identity sin2A + cos2A = 1,
= (sin θ + cos θ)/sin θ cos θ
= (sin θ/sin θ cos θ) + (cos θ/sin θ cos θ)
= (1/cos θ) + (1/sin θ)
= sec θ + cosec θ
= RHS
Hence proved.
Question 15: Prove that: sin2θ cos θ + tan θ sin θ + cos3θ = sec θ
Solution:
LHS = sin2θ cos θ + tan θ sin θ + cos3θ
= (sin2θ cos θ + cos3θ) + tan θ sin θ
= cos θ(sin2θ + cos2θ) + tan θ sin θ
Using the identity sin2A + cos2A = 1,
= cos θ(1) + (sin θ/cos θ) sin θ
= cos θ + (sin2θ/cos θ)
= (cos2θ + sin2θ)/cos θ
= 1/cos θ
= sec θ
= RHS
Hence proved.
Question 16: Prove that: sin8θ – cos8θ = (sin2θ – cos2θ) (1 – 2 sin2θ cos2θ) using the identities.
Solution:
LHS = sin8θ – cos8θ
= (sin4θ)2 – (cos4θ)2
= (sin4θ + cos4θ) (sin4θ – cos4θ)
= [(sin2θ)2 + (cos2θ)2 + 2 cos2θ sin2θ – 2 sin2θ cos2θ] [(sin2θ)2 – (cos2θ)2]
= [(sin2θ + cos2θ)2 – 2 sin2θ cos2θ] [(sin2θ + cos2θ) (sin2θ – cos2θ)]
Using the identity sin2A + cos2A = 1,
= (1 – 2 sin2θ cos2θ) (sin2θ – cos2θ)
= (sin2θ – cos2θ) (1 – 2 sin2θ cos2θ)
=RHS
Hence proved.
Question 17: Prove that: [(1 + cot A + tan A)(sin A – cos A) / (sec3A – cosec3A)] = sin2A cos2A
Solution:
LHS = [(1 + cot A + tan A)(sin A – cos A) / (sec3A – cosec3A)]
= [(1 + cos A/sin A + sin A/cos A)(sin A – cos A)] / [(1/cos3A) – (1/sin3A)]
= [(sin A cos A + cos2A + sin2A)(sin A – cos A)/sin A cos A] / [(sin3A – cos3A)/ sin3A cos3A]
= [(sin3A – cos3A) sin3A cos3A] / [sin A cos A(sin3A – cos3A)]
= (sin3A cos3A)/(sin A cos A)
= sin2A cos2A
= RHS
Hence proved.
Question 18: Prove that: [(sin θ + cos θ)/(sin θ – cos θ)] + [(sin θ – cos θ)/(sin θ + cos θ)] = [2/(1 – 2 cos2θ)] = [2/ (2 sin2θ – 1)]
Solution:
LHS = [(sin θ + cos θ)/(sin θ – cos θ)] + [(sin θ – cos θ)/(sin θ + cos θ)]
= [(sin θ + cos θ)2 + (sin θ – cos θ)2] / [(sin θ – cos θ)(sin θ + cos θ)]
= [sin2θ + cos2θ + 2 sin θ cos θ + sin2θ + cos2θ – 2 sin θ cos θ] / (sin2θ – cos2θ)
Using the identity sin2A + cos2A = 1,
= (1 + 1)/ (sin2θ – cos2θ)
= 2/[(1 – cos2θ) – cos2θ]
= 2/(1 – 2 cos2θ) ….(i)
Now,
2/(sin2θ – cos2θ) = 2/[sin2θ – (1 – sin2θ)]
= 2/(2 sin2θ – 1) ….(ii)
From (i) and (ii),
[(sin θ + cos θ)/(sin θ – cos θ)] + [(sin θ – cos θ)/(sin θ + cos θ)] = [2/(1 – 2 cos2θ)] = [2/ (2 sin2θ – 1)]Hence proved.
Question 19: Prove that: [cos A/(1 – tan A)] + [sin A/(1 – cot A)] = sin A + cos A
Solution:
[cos A/(1 – tan A)] = cos A/[1 – (sin A/cos A)]= cos A/ [(cos A – sin A)/cos A]
= cos2A/(cos A – sin A)
Now,
[sin A/(1 – cot A)] = sin A/ [1 – (cos A/sin A)]= sin A/[(sin A – cos A)/sin A]
= sin2A/(sin A – cos A)
LHS = [cos A/(1 – tan A)] + [sin A/(1 – cot A)]
= cos2A/(cos A – sin A) + sin2A/(sin A – cos A)
= sin2A/(sin A – cos A) – cos2A/(sin A – cos A)
= (sin2A – cos2A)/(sin A – cos A)
= [(sin A + cos A)(sin A – cos A)]/ (sin A – cos A)
= sin A + cos A
= RHS
Hence proved.
Question 20: Prove that: [cos2θ/(1 – tan θ)] + [sin3θ/(sin θ – cos θ)] = 1 + sin θ cos θ
Solution:
LHS = [cos2θ/(1 – tan θ)] + [sin3θ/(sin θ – cos θ)]
= [cos2θ/(1 – sin θ/cosθ)] + [sin3θ/(sin θ – cos θ)]
= [cos3θ/(cos θ – sin θ)] + [sin3θ/(sin θ – cos θ)]
= [sin3θ/(sin θ – cos θ)] – [cos3θ/(sin θ – cos θ)]
= (sin3θ – cos3θ)/(sin θ – cos θ)
= [(sin θ – cos θ)(sin2θ + cos2θ + sin θ cos θ]/ (sin θ – cos θ)
Using the identity sin2A + cos2A = 1,
= 1 + sin θ cos θ
= RHS
Hence proved.
Question 21: If sec θ + tan θ = p, then prove that: (p2 – 1)/(p2 + 1) = sin θ
Solution:
Given,
sec θ + tan θ = p
LHS = (p2 – 1)/(p2 + 1)
= [(sec θ + tan θ)2 – 1] / [(sec θ + tan θ)2 + 1]
= [sec2θ + tan2θ + 2 sec θ tan θ – 1] / [sec2θ + tan2θ + 2 sec θ tan θ + 1]
= [(sec2θ – 1) + (tan2θ + 2 sec θ tan θ)] / [(sec2θ + tan2θ + 1) + 2 sec θ tan θ]
Using the identity 1 + tan2A = sec2A,
= [tan2θ + tan2θ + 2 sec θ tan θ] / [sec2θ + sec2θ + 2 sec θ tan θ]
= [2 tan2θ + 2 sec θ tan θ] / [2 sec2θ + 2 sec θ tan θ]
= [2 tan θ(tan θ + sec θ)] / [2 sec θ(tan θ + sec θ)]
= tan θ/sec θ
= (sin θ/cos θ) × cos θ
= sin θ
= RHS
Hence proved.
Question 22: If (cos A/cos B) = m and (cos A/sin B) = n, then prove that (m2 + n2)cos2B = n2.
Solution:
Given,
(cos A/cos B) = m and (cos A/sin B) = n
LHS = (m2 + n2)cos2B
= [(cos2A/cos2B) + (cos2A/sin2B)] cos2B
= [(cos2A sin2B + cos2A cos2B)/ (cos2B sin2B)] cos2B
= [cos2A(sin2B + cos2B)] / sin2B
Using the identity sin2θ + cos2θ = 1,
= cos2A/sin2B
= (cos A/sin B)2
= n2 [from the given)
= RHS
Hence proved.
RBSE Maths Chapter 7: Exercise 7.2 Textbook Important Questions and Solutions
Question 23: Find the value of cos 37°/sin 53°.
Solution:
cos 37°/sin 53° = cos(90° – 53°)/sin 53°
We know that cos(90° – A) = sin A
= sin 53°/sin 53°
= 1
Question 24: Find the value of cosec 32°/sec 58°.
Solution:
cosec 32°/sec 58° = cosec(90° – 58°)/sec 58°
We know that cosec(90° – A) = sec A
= sec 58°/sec 58°
= 1
Question 25: Find the value of tan 10°/cot 80°.
Solution:
tan 10°/cot 80° = tan(90° – 80°)/cot 80°
We know that tan(90° – A) = cot A
= cot 80°/cot 80°
= 1
Question 26: Find the value of cosec 25° – sec 65°.
Solution:
cosec 25° – sec 65°
= cosec(90° – 65°) – sec 65°
We know that cosec(90° – θ) = sec θ
= sec 65° – sec 65°
= 0
Question 27: Find the value of (sin 36°/cos 54°) – (sin 54°/cos 36°).
Solution:
(sin 36°/cos 54°) – (sin 54°/cos 36°)
= [sin(90° – 54°)/cos 54°] – [sin(90° – 36°)/cos 36°]
We know that sin(90° – A) = cos A
= (cos 54°/cos 54°) – (cos 36°/cos 36°)
= 1 – 1
= 0
Question 28: Find the value of sin θ cos(90° – θ) + cos θ sin(90° – θ).
Solution:
sin θ cos(90° – θ) + cos θ sin(90° – θ)
We know that cos (90° – A) = sin A and sin (90° – A) = cos A
= sin θ sin θ + cos θ cos θ
= sin2θ + cos2θ
Using the identity sin2A + cos2A = 1,
= 1
Question 29: Find the value of sin 70° sin 20° – cos 20° cos 70°.
Solution:
sin 70° sin 20° – cos 20° cos 70°
= sin(90° – 20°) sin 20° – cos 20° cos(90° – 20°)
We know that cos (90° – A) = sin A and sin (90° – A) = cos A
= cos 20° sin 20° – cos 20° sin 20°
= 0
Question 30: Find the value of (2 cos 67°/sin 23°) – (tan 40°/cot 50°) – cos 60°.
Solution:
(2 cos 67°/sin 23°) – (tan 40°/cot 50°) – cos 60°
= [2 cos(90° – 23°)/sin 23°] – [tan(90° – 50°)/cot 50°) – (1/2)
We know that cos(90° – A) = sin A and tan(90° – A) = cot A
= (2 sin 23°/sin 23°) – (cot 50°/cot 50°) – (1/2)
= 2(1) – 1 – (1/2)
= 1 – ½
= (2 – 1)/2
= 1/2
Question 31: Find the value of (sin 35°/cos 55°)2 + (cos 55°/sin 35°)2 – 2 cos 60°.
Solution:
(sin 35°/cos 55°)2 + (cos 55°/sin 35°)2 – 2 cos 60°
= [sin(90° – 55°)/cos 55°]2 + [cos(90° – 35°)/sin 35°]2 – 2 × (1/2)
We know that sin(90° – A) = cos A and cos(90° – A) = sin A
= (cos 55°/cos 55°)2 + (sin 35°/sin 35°)2 – 1
= (1)2 + (1)2 – 1
= 1 + 1 – 1
= 1
Question 32: Find the value of tan 5° tan 25° tan 30° tan 45° tan 65° tan 85°.
Solution:
tan 5° tan 25° tan 30° tan 45° tan 65° tan 85°
= tan 5° tan 25° tan 30° × 1 × tan(90° – 25°) tan(90° – 5°)
We know that tan(90° – θ) = cot θ
= tan 5° × tan 25° × tan 30° × cot 25° × cot 5°
= tan 5° × (1/tan 5°) × tan 25° × (1/tan 25°) × tan 30°
= 1 × 1 × (1/√3)
= 1/√3
Question 33: Express the following in terms of trigonometric ratios of angles between 0° and 45°.
(i) sin 81° + sin 71°
(ii) tan 68° + sec 68°
Solution:
(i) sin 81° + sin 71°
= sin (90° – 9°) + sin (90° – 19°)
We know that sin (90° – A) = cos A
= cos 9° + cos 19°
(ii) tan 68° + sec 68°
= tan(90° – 22°) + sec(90° – 22°)
We know that tan(90° – A) = cot A and sec(90° – A) = cosec A
= cot 22° + cosec 22°
Question 34: Prove that: sin 65° + cos 25° = 2 cos 25°
Solution:
LHS = sin 65° + cos 25°
= sin(90° – 25°) + cos 25°
We know that sin(90° – A) = cos A
= cos 25° + cos 25°
= 2 cos 25°
= RHS
Hence proved.
Question 35: Prove that: (cos 70°/sin 20°) + (cos 59°/sin 31°) – 8 sin230° = 0
Solution:
LHS = (cos 70°/sin 20°) + (cos 59°/sin 31°) – 8 sin230°
= [cos(90° – 20°)/sin 20°] + [cos(90° – 31°)/sin 31°] – 8 × (1/2)2
We know that cos(90° – A) = sin A
= (sin 20°/sin 20°) + (sin 31°/sin 31°) – 8 × (1/4)
= 1 + 1 – 2
= 0
= RHS
Hence proved.
Question 36: Prove that: sin(90° – θ) cos(90° – θ) = tan θ/(1 + tan2θ)
Solution:
sin(90° – θ) cos(90° – θ)
= cos θ sin θ
Multiplying and dividing by cos θ,
= (sin θ cos θ cos θ)/cos θ
= (sin θ/cos θ) × cos2θ
= tan θ × (1/sec2θ)
Using the identity 1 + tan2A = sec2A,
= tan θ/(1 + tan2θ)
= RHS
Hence proved.
Question 37: Prove that: [cos(90° – θ) cos θ/tan θ] + cos2(90° – θ) = 1
Solution:
LHS = [cos(90° – θ) cos θ/tan θ] + cos2(90° – θ)
We know that cos(90° – A) = sin A
= [sin θ cos θ/tan θ] + sin2θ
= [sin θ × cos θ × cos θ/sin θ] + sin2θ
= cos2θ + sin2θ
Using the identity sin2A + cos2A = 1,
= 1
= RHS
Hence proved.
Question 38: Prove that: [tan(90° – θ) cot θ/cosec2θ] – cos2θ = 0
Solution:
LHS = [tan(90° – θ) cot θ/cosec2θ] – cos2θ
= cot θ × cot θ × sin2θ – cos2θ
= cot2θ × sin2θ – cos2θ
= (cos2θ/sin2θ) × sin2θ – cos2θ
= cos2θ – cos2θ
= 0
= RHS
Hence proved.
Question 39: Prove that: [sin θ cos(90° – θ) cos θ/sec(90° – θ)] + [cos θ sin(90° – θ) sin θ/cosec(90° – θ)] = sin θ cos θ
Solution:
LHS = [sin θ cos(90° – θ) cos θ/sec(90° – θ)] + [cos θ sin(90° – θ) sin θ/cosec(90° – θ)]
= [sin θ × sin θ × cos /cosec θ] + [cos θ × cos θ × sin θ /sec θ]
= (sin2θ × cosθ × sin θ) + (cos2θ × sin θ × cos θ)
= (sin3θ × cos θ) + (cos3θ × sin θ)
= sin θ cos θ(sin2θ + cos2θ)
Using the identity sin2A + cos2A = 1,
= (sin θ cos θ) × (1)
= sin θ cos θ
= RHS
Hence proved.
Question 40: If sin 3θ = cos(θ – 6°), where 3θ and (θ – 6°) are acute angles, then find the value of θ.
Solution:
Given,
sin 3θ = cos(θ – 6°), where 3θ and (θ – 6°) are acute angles
We know that cos(90° – A) = sin A
Thus, cos(90° – 3θ) = cos(θ – 6°)
⇒ 90° – 3θ = θ – 6°
⇒ 3θ + θ = 90° + 6°
⇒ 4θ = 96°
⇒ θ = 96°/4
⇒ θ = 24°
Question 41: If A, B, and C are interior angles of a triangle ABC, then prove that: tan(B + C/2) = cot(A/2)
Solution:
We know that, sum of all the interior angles of a triangle is equal to 180°.
Thus, A + B + C = 180°
B + C = 180° – A
⇒ (B + C)/2 = (180° – A)/2
⇒ (B + C)/2 = (90° – A/2)
Now, multiply both sides by tan functions,
⇒ tan(B + C/2) = tan(90°-A/2)
We know that tan(90° – θ) = cot θ,
tan(B + C/2) = cot(A/2)
Hence proved.
RBSE Maths Chapter 7: Additional Important Questions and Solutions
Question 1: If sin 2A = cos(A – 18°), then find the value of A.
Solution:
Given,
sin 2A = cos(A – 18°)
We know that cos(90° – θ) = sin θ
⇒ cos(90° – 2A) = cos(A – 18°)
⇒ 90° – 2A = A – 18°
⇒ 2A + A = 90° + 18°
⇒ 3A = 108°
⇒ A = 108°/3
⇒ A = 36°
Question 2: Write the value of cos 50°. cosec 40°.
Solution:
cos 50° cosec 40°
= cos(90° – 40°) cosec 40°
We know that cos(90° – A) = sin A
= sin 40° × (1/sin 40°)
= 1
Question 3: Prove that: √[(1 + cos A)/(1 – cos A)] = cosec A + cot A
Solution:
LHS = √[(1 + cos A)/(1 – cos A)]
By rationalising the denominator,
√[1 + cos A)2/(1 – cos2A)]
Using the identity sin2θ + cos2θ = 1,
= √[(1 + cos A)2/sin2A]
= (1 + cos A)/sin A
= (1/sin A) + (cos A/sin A)
= cosec A + cot A
= RHS
Hence proved.
Question 4: If sin θ + cos θ = p, sec θ + cosec θ = q, then prove that: q(p2 – 1) = 2p.
Solution:
Given,
sin θ + cos θ = p and sec θ + cosec θ = q
LHS = q(p2 – 1)
= (sec θ + cosec θ)[(sin θ + cos θ)2 – 1]
= (sec θ + cosec θ) [sin2θ + cos2θ + 2 sin θ cos θ – 1]
Using the identity sin2A + cos2A = 1,
= (sec θ + cosec θ) [1 + 2 sin θ cos θ – 1]
= [(1/cos θ) + (1/sin θ)] × (2 sin θ cos θ)
= [(sin θ + cos θ)/sin θ cos θ] × (2 sin θ cos θ)
= 2(sin θ + cos θ)
= 2p
= RHS
Hence proved.
Question 5: Write the trigonometric ratio of sin A in terms of cot A.
Solution:
We know that,
cot²A + 1 = cosec²A
⇒ cosec A = √(1 + cot²A)
sin A = 1/cosec A
= 1/√(1 + cot²A)
Question 6: Prove that: (tan A – sin A)/(tan A + sin A) = (sec A – 1)/(sec A + 1)
Solution:
LHS = (tan A – sin A) / (tan A + sin A)
= [(sin A/cos A) – sin A] / [(sin A/cos A) + sin A]
= [(sin A – sin A cos A)/cos A] / [(sin A + sin A cos A)/ cos A]
= [sin A(1 – cos A)/cos A] × [cos A/ sin A(1 + cos A)]
= (1 – cos A)/(1 + cos A)
Dividing the numerator and denominator by cos A,
= [(1/cos A) – (cos A/cos A)] / [(1/cos A) + (cos A/cos A)]
= (sec A – 1)/(sec A + 1)
= RHS
Hence proved.
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