RBSE Maths Chapter 14 – Trigonometric Ratios of Acute Angles Class 9 Important questions and solutions are available here. The important questions and solutions of Chapter 14 provided at BYJU’S have detailed steps. Also, RBSE Class 9 solutions are given at BYJU’S to help the students in getting maximum marks in their exams and in improving their problem-solving skills.
Chapter 14 of the RBSE Class 9 Maths has formulas and identities of trigonometric ratios. Students will be introduced to finding the values of all the six trigonometric functions using a right triangle. Also, there exist problems in proving the trigonometric expressions using the relevant identities of trigonometry.
RBSE Maths Chapter 14: Exercise 14.1 Textbook Important Questions and Solutions
Question 1: If in a ΔABC, ∠A = 90°, a = 25 cm, b = 7 cm, then find all the trigonometric ratios of ∠B and ∠C.
Solution:
Given,
∠A = 90°, a = 25 cm, b = 7 cm
Thus, BC = 25 cm and CA = 7 cm
In right triangle ABC,
AB2 + AC2 = BC2
AB2 = BC2 – AC2
AB = √(25)2 – (7)2
= √(625 – 49)
= √576
AB = 24
For acute angle B, trigonometric ratios are:
sin B = AC/BC = 7/25
cos B = AB/BC = 24/25
tan B = AC/AB = 7/24
cosec B = 1/sin B = 25/7
sec B = 1/cos B = 25/24
cot B = 1/tan B = 24/7
For acute angle C, trigonometric ratios are:
sin C = AB/BC = 24/25
cos C = AC/BC = 7/25
tan C = AB/AC = 24/7
cosec C = 1/sin C = 25/24
sec C = 1/cos C = 25/7
cot C = 1/tan C = 7/24
Question 2: If in a ΔABC, ∠B = 90°, a = 12 cm, b = 13 cm, then find all the trigonometric ratios of ∠A and ∠C.
Solution:
Given,
∠B = 90°, a = 12 cm, b = 13 cm
Thus, BC = 12 cm and AC = 13 cm
In right triangle ABC,
AB2 + BC2 = AC2
AB2 = AC2 – BC2
= 132 – 122
= 169 – 144
= 25
AB = √25 = 5 cm
For acute angle A, all trigonometric ratios are:
sin A = BC/AC = 12/13
cos A = AB/AC = 5/13
tan A = BC/AB = 12/5
cosec A = 1/sin A = 13/12
sec A = 1/cos A = 13/5
cot A = 1/tan A = 5/12
For acute angle C, all trigonometric ratios are:cot A = 1/tan A = 5/12
sin C = AB/AC = 5/13
cos C = BC/AC = 12/13
tan C = AB/BC = 5/12
cosec C = 1/sin C = 13/5
sec C = 1/cos C = 13/12
cot C = 1/tan C = 12/5
Question 3: If tan A = √2 – 1, then prove that sin A cos A = 1/2√2.
Solution:
Given,
tan A = √2 – 1
Construct a triangle ABC in which ∠B = 90° and BC/AB = (√2 – 1)/1.
Let BC = (√2 – 1)k and AB = k, where k is proportionality constant.
In right triangle ABC,
AC2 = AB2 + BC2
= k2 + [(√2 – 1)k]2
= k2 + (2 – 2√2 + 1)k2
= k2 – 2√2k2 + 3k2
= (4 – 2√2)k2
AC = [√(4 – 2√2)]k
sin A = BC/AC = (√2 – 1)k/[√(4 – 2√2)]k = (√2 – 1)/√(4 – 2√2)
cos A = AB/AC = k/[(4 – 2√2)]k = 1/√(4 – 2√2)
LHS = sin A cos A
= [(√2 – 1)/√(4 – 2√2)] × [1/√(4 – 2√2)]
= (√2 – 1)/ (4 – 2√2)
= [(√2 – 1)/(4 – 2√2)] × [(4 + 2√2)/ (4 + 2√2)]
= [(√2 – 1) (4 + 2√2)}/ [(4)2 – (2√2)2]
= (4√2 + 4 – 4 – 2√2)/(16 – 8)
= (2√2)/8
= (2√2)/ (2√2)2
= 1/2√2
= RHS
Hence proved.
Question 4: If sin A = 1/3, then find the value of cos A cosec A + tan A sec A.
Solution:
Given,
sin A = ⅓ = Perpendicular/Hypotenuse
Construct a right triangle ABC in which, B is a right angle, BC = k, AC = 3k
In right triangle ABC,
AB2 + BC2 = AC2
AB2 = AC2 – BC2
= (3k)2 – k2
= 9k2 – k2
= 8k2
AB = √(8k2) = 2√2 k
cos A = AB/AC = 2√2k/3k = (2√2)/3
tan A = BC/AB = k/(2√2k) = 1/2√2
cosec A = 1/sin A = 3
sec A = 1/cos A = 3/2√2
cos A cosec A + tan A sec A = [(2√2)/3] × 3 + [1/(2√2)] × [3/(2√2)]
= 2√2 + [3/ (2√2)2]
= 2√2 + (⅜)
Question 5: If cos θ = 8/17, then find all the remaining trigonometric ratios.
Solution:
Given,
cos θ = 8/17 = Base/Hypotenuse
Construct a right triangle ABC in which, B is a right angle, AC = 17 k and AB = 8k.
In right triangle ABC,
AB2 + BC2 = AC2
BC2 = AC2 – AB2
= (17k)2 – (8k)2
= 289k2 – 64k2
= 225k2
BC = √(225k2) = 15k
sin θ = BC/AC = 15k/17k = 15/17
tan θ = BC/AB = 15/8
cosec θ = 1/sin θ = 17/15
sec θ = 1/cos θ = 17/8
cot θ = 1/tan θ = 8/15
Question 6: If cos A = 5/13, then find the value of cosec A/(cos A + cosec A).
Solution:
Given,
cos A = 5/13
Construct a right triangle ABC in which, B is a right angle, AC = 13k and AB = 5k.
In right triangle ABC,
AB2 + BC2 = AC2
BC2 = AC2 – AB2
= (13k)2 – (5k)2
= 169k2 – 25k2
= 144k2
BC = √(144k2) = 12k
cosec A = AC/BC = 13k/12k = 13/12
cosec A/ (cos A + cosec A) = (13/12)/ [(5/13) + (13/12)]
= (13/12)[(60 + 169)/156]
= (13/12) × (156/229)
= 169/229
Question 7: If 5 tan θ = 4, then find the value of (5 sin θ – 3 cos θ)/(sin θ + 2 cos θ).
Solution:
Given,
5 tan θ = 4
tan θ = ⅘ = Perpendicular/Base
Construct a triangle ABC in which B is a right angle and AB = 5k and BC = 4k
AC2 = AB2 + BC2
= (5k)2 + (4k)2
= 25k2 + 16k2
= 41k2
AC = √(41k2) = √41 k
sin θ = BC/AC = 4k/√41 k = 4/√41
cos θ = AB/AC = 5k/√41 k = 5/√41
(5 sin θ – 3 cos θ)/ (sin θ + 2 cos θ) = [5(4/√41) – 3(5/√41)]/[(4/√41) + 2(5/√41)]
= [(20 – 15)/√41] × [(√41)/(4 + 10)]
= 5/14
Question 8: In ΔABC, ∠C = 90° and if cot A = √3 and cot B = 1/√3, then prove that sin A cos B + cos A sin B = 1.
Solution:
Given,
cot A = √3 and cot B = 1/√3
cot A = √3/1 = Base/Perpendicular
Construct a right triangle ABC in which B is right angle, AC =√3k and BC = k
AB2 = AC2 + BC2
= (√3k)2 + (k)2
= 3k2 + k2
= 4k2
AC = √(4k2) = 2k
sin A = BC/AB = k/2k = ½
cos A = AC/AB = √3k/2k = √3/2
sin B = AC/AB = √3k/2k = √3/2
cos B = BC/AB = k/2k = ½
LHS = sin A cos B + cos A sin B
= (½)(1/2) + (√3/2)(√3/2)
= (¼) + (¾)
= (1 + 3)/4
= 4/4
= 1 = RHS
Hence proved.
Question 9: If 16 cot A = 12, then find the value of (sin A + cos A)/(sin A – cos A).
Solution:
Given,
16 cot A = 12
cot A = 12/16 = 3/4
(sin A + cos A)/(sin A – cos A)
Dividing the numerator and denominator by sin A
= [(sin A/sin A) + (cos A/sin A)]/ [(sin A/sin A) – (cos A/sin A)]
= (1 + cot A)/(1 – cot A)
= [1 + (3/4)]/ [1 – (3/4)]
= [(4 + 3)/4]/ [(4 – 3)/4]
= 7
RBSE Maths Chapter 14: Exercise 14.2 Textbook Important Questions and Solutions
Solve the following with the help of relation between trigonometric ratios.
Question 10: If cosec A = 5/4, then find the value of cot A, sin A and cos A.
Solution:
Given,
cosec A = 5/4
We know that,
cot2A + 1 = cosec2A
cot2A = cosec2A – 1
= (5/4)2 – 1
= (25/16) – 1
= (25 – 16)/16
= 9/16
cot A = √(9/16) = 3/4
sin A = 1/cosec A = 4/5
Now,
cot A = cos A/sin A
cos A = sin A cot A
= (4/5) × (3/4)
= 3/5
Therefore, cot A = 3/4, sin A = 4/5 and cos A = 3/5
Question 11: If cos B = 1/3, then find remaining trigonometric ratios.
Solution:
Given,
cos B = ⅓
We know that,
sin2B + cos2B = 1
sin2B = 1 – cos2B
= 1 – (1/3)2
= 1- (1/9)
= (9 – 1)/9
= 8/9
sin B = √(8/9) = 2√2/3
tan B = sin B/cos B
= (2√2/3)/ (1/3)
= 2√2
cosec B = 1/sin B = 3/2√2
sec B = 1/cos B = 3
cot B = 1/tan B = 1/2√2
Therefore, the remaining trigonometric ratios are:
sin B = 2√2/3, tan B = 2√2, cosec B = 3/2√2, sec B = 3, cot B = 1/2√2
Question 12: If sin A = 5/13, then find the value of cos A and tan A.
Solution:
Given,
sin A = 5/13
We know that,
sin2A + cos2A = 1
cos2A = 1 – sin2A
= 1 – (5/13)2
= 1- (25/169)
= (169 – 25)/169
= 144/169
cos A = √(144/169) = 12/13
tan A = sin A/cos A
= (5/13)/ (12/13)
= 5/12
Therefore, cos A = 12/13 and tan A = 5/12
Question 13: If tan A = 2, then find the value of sec A sin A + tan2A – cosec A.
Solution:
Given,
tan A = 2
We know that,
sec2A = 1 + tan2A
= 1 + (2)2
= 1 + 4
= 5
sec A = √5
cos A = 1/sec A = 1/√5
Now,
tan A = sin A/cos A
sin A = cos A tan A
= (1/√5) × (2)
= 2/√5
cosec A = 1/sin A = √5/2
sec A sin A + tan2A – cosec A = (√5) × (2/√5) + (2)2 – (√5/2)
= 2 + 4 – √5/2
= 6 – √5/2
= (12 – √5)/2
Question 14: If sin θ = 4/5, then find the value of (4 tan θ – 5 cos θ)/(sec θ + 4 cot θ).
Solution:
Given,
sin θ = 4/5
We know that,
sin2θ + cos2θ = 1
cos2θ = 1 – sin2θ
= 1 – (4/5)2
= 1 – (16/25)
= (25 – 16)/25
= 9/25
cos θ = √(9/25) = 3/5
tan θ = sin θ/cos θ
= (4/5)/ (3/5)
= 4/3
cot θ = 1/tan θ = 3/4
sec θ = 1/cos θ = 5/3
Now,
(4 tan θ – 5 cos θ)/ (sec θ + 4 cot θ) = [4 × (4/3) – 5 × (3/5)] / [(5/3) + 4 × (3/4)]
= [(16/3) – 3] / [(5/3) + 3]
= (16 – 9)/ (5 + 9)
= 7/14
= 1/2
Question 15: If sec θ = 2, then evaluate tan θ, cos θ and sin θ.
Solution:
Given,
sec θ = 2
We know that,
sec2θ = 1 + tan2θ
tan2θ = sec2θ – 1
= (2)2 – 1
= 4 – 1
= 3
tan θ = √3
cos θ = 1/sec θ = 1/2
Now,
tan θ = sin θ/cos θ
sin θ = tan θ cos θ
= (√3) × (1/2)
= √3/2
Therefore, tan θ = √3, cos θ = 1/2, and sin θ = √3/2
RBSE Maths Chapter 14: Exercise 14.3 Textbook Important Questions and Solutions
Question 16: Prove that: cos θ.tan θ = sin θ
Solution:
LHS = cos θ.tan θ
= cos θ × (sin θ/cos θ)
= sin θ
= RHS
Therefore, cos θ.tan θ = sin θ
Hence proved.
Question 17: Prove that: (1 – sin2θ) tan2θ = sin2θ
Solution:
LHS = (1 – sin2θ) tan2θ
Using the identity sin2θ + cos2θ = 1,
= cos2θ × (sin2θ/cos2θ)
= sin2θ
= RHS
Therefore, (1 – sin2θ) tan2θ = sin2θ
Hence proved.
Question 18: Prove that: (cos2θ/sin θ) + sin θ = cosec θ
Solution:
LHS = (cos2θ/sin θ) + sin θ
= (cos2θ + sin2θ)/sin θ
Using the identity sin2θ + cos2θ = 1,
= 1/sin θ
= cosec θ
= RHS
Therefore, (cos2θ/sin θ) + sin θ = cosec θ
Hence proved.
Question 19: Prove that: (sin θ + cos θ)2 + (sin θ – cos θ)2 = 2
Solution:
LHS = (sin θ + cos θ)2 + (sin θ – cos θ)2
= sin2θ + cos2θ + 2 sin θ cos θ + sin2θ + cos2θ – 2 sin θ cos θ
= 2 sin2θ + 2 cos2θ
= 2(sin2θ + cos2θ)
Using the identity sin2θ + cos2θ = 1,
= 2(1)
= 2
= RHS
Therefore, (sin θ + cos θ)2 + (sin θ – cos θ)2 = 2
Hence proved.
Question 20: Prove that: cosec6θ – cot6θ = 1 + 3 cosec2θ cot2θ
Solution:
LHS = cosec6θ – cot6θ
= (cosec2θ)3 – (cot2θ)3
= (cosec2θ – cot2θ) [(cosec2θ)2 + cosec2θ cot2θ + (cot2θ)2]
Using the identity cot2θ + 1 = cosec2θ,
= (1)[cosec4θ + cot4θ + cosec2θ cot2θ]
= (cosec2θ – cot2θ)2 + 2 cosec2θ cot2θ + cosec2θ cot2θ
= (1)2 + 3 cosec2θ cot2θ
= 1 + 3 cosec2θ cot2θ
= RHS
Thus, cosec6θ – cot6θ = 1 + 3 cosec2θ cot2θ
Hence proved.
Question 21: 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) + (sin θ/cos θ) × sin θ
Using the identity sin2θ + cos2θ = 1,
= cos θ × (1) + (sin2θ/cos θ)
= cos θ + (sin2θ/cos θ)
= (cos2θ + sin2θ)/cos θ
= 1/cos θ
= sec θ
= RHS
Thus, sin2θ cos θ + tan θ sin θ + cos3θ = sec θ
Hence proved.
Question 22: Prove that: [cos θ/ (1 – tan θ)] + [sin θ/ (1 – cot θ)] = sin θ + cos θ
Solution:
LHS = [cos θ/(1 – tan θ)] + [sin θ/(1 – cot θ)]
= [cos θ/ (1 – sin θ/cos θ)] + [sin θ/ (1 – cos θ/sin θ)]
= [cos2θ / (cos θ – sin θ)] + [sin2θ / (sin θ – cos θ)]
= [cos2θ / (cos θ – sin θ)] – [sin2θ / (cos θ – sin θ)]
= (cos2θ – sin2θ)/(cos θ – sin θ)
= [(cos θ + sin θ)(cos θ – sin θ)]/(cos θ – sin θ)
= cos θ + sin θ
= RHS
Thus, [cos θ/ (1 – tan θ)] + [sin θ/ (1 – cot θ)] = sin θ + cos θ
Hence proved.
Question 23: Prove that: [cosec θ/ (cosec θ – 1)] + [cosec θ/ (cosec θ + 1] = 2 sec2θ
Solution:
LHS = [cosec θ/ (cosec θ – 1)] + [cosec θ/ (cosec θ + 1]
= cosec θ [1/ (cosec θ – 1)] + [1/ (cosec θ + 1]
= cosec θ [(cosec θ + 1 + cosec θ – 1) / (cosec θ – 1)(cosec θ + 1)]
= cosec θ [2 cosec θ/ (cosec2θ – 12)]
= 2 cosec2θ/ (cosec2θ – 1)
Using the identity cot2θ + 1 = cosec2θ,
= 2 cosec2θ/cot2θ
= 2 × cosec2θ × tan2θ
= 2 × (1/ sin2θ) × (sin2θ/cos2θ)
= 2/cos2θ
= 2 sec2θ
= RHS
Thus, [cosec θ/ (cosec θ – 1)] + [cosec θ/ (cosec θ + 1] = 2 sec2θ
Hence proved.
Question 24: Prove that: sin θ/(1 – cos θ) = (1 + cos θ)/sin θ
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)/ (12 – cos2θ)
= sin θ (1 + cos θ)/ (1 – cos2θ)
Using the identity sin2θ + cos2θ = 1,
= sin θ(1 + cos θ)/sin2θ
= (1 + cos θ)/sin θ
= RHS
Thus, sin θ/(1 – cos θ) = (1 + cos θ)/sin θ
Hence proved.
Question 25: Prove that: [sin θ/ (1 + cos θ)] + [(1 + cos θ)/sin θ] = 2 cosec θ
Solution:
LHS = [sin θ/ (1 + cos θ)] + [(1 + cos θ)/sin θ]
= [sin2θ + (1 + cos θ)2]/(1 + cos θ)sin θ
= [sin2θ + 1 + cos2θ + 2 cos θ]/(1 + cos θ) sin θ
Using the identity sin2θ + cos2θ = 1,
= (1 + 1 + 2 cos θ)/(1 + cos θ) sin θ
= (2 + 2 cos θ)/(1 + cos θ) sin θ
= 2(1 + cos θ)/(1 + cos θ) sin θ
= 2/sin θ
= 2 cosec θ
= RHS
Thus, [sin θ/ (1 + cos θ)] + [(1 + cos θ)/sin θ] = 2 cosec θ
Hence proved.
Question 26: Prove that:
Solution:
By rationalising the denominator,
Thus,
Hence proved.
Question 27: Prove that:
Solution:
By rationalising the denominator,
= sec θ cot θ + cot θ
= (1/cos θ) (cos θ/sin θ) + cot θ
= cosec θ + cot θ
= RHS
Thus,
Hence proved.
Question 28: Prove that: [√(cosec2θ – 1)]/cosec θ = cos θ
Solution:
LHS = [√(cosec2θ – 1)]/cosec θ
Using the identity cot2θ + 1 =cosec2θ,
= √(cot2θ)/cosec θ
= cot θ/cosec θ
= cot θ × sin θ
= (cos θ/sin θ) × sin θ
= cos θ
= RHS
Thus, [√(cosec2θ – 1)]/cosec θ = cos θ
Hence proved.
Question 29: Prove that: (1 + cot θ – cosec θ)(1 + tan θ + sec θ) = 2
Solution:
LHS = (1 + cot θ – cosec θ)(1 + tan θ + sec θ)
= [1 + (cos θ/sin θ) – (1/sin θ)] [1 + (sin θ/cos θ) + (1/cos θ)]
= [(sin θ + cos θ – 1)/ sin θ] [(cos θ + sin θ + 1)/ cos θ]
= [(sin θ + cos θ)2 – 12] / sin θ cos θ
= (sin2θ + cos2θ + 2 sin θ cos θ – 1)/ sin θ cos θ
Using the identity sin2θ + cos2θ = 1,
= (1 + 2 sin θ cos θ – 1)/sin θ cos θ
= (2 sin θ cos θ)/sin θ cos θ
= 2
= RHS
Thus, (1 + cot θ – cosec θ)(1 + tan θ + sec θ) = 2
Hence proved.
RBSE Maths Chapter 14: Additional Important Questions and Solutions
Question 1: If √3 cos A = sin A, then the value of cot A is:
(A) √3
(B) 1
(C) 1/√3
(D) 2
Solution:
Correct answer: (C)
Given,
√3 cos A = sin A
cos A/sin A = 1/√3
cot A = 1/√3
Question 2: In given ΔABC, the value of cosec α is:
(A) y/x
(B) y/z
(C) x/z
(D) x/y
Solution:
Correct answer: (B)
Given,
AB = z, BC = x and AC = y
cosec α = Hypotenuse / Opposite side to α
= AC/AB
= y/z
Therefore, cosec α = y/z
Question 3: The value of sin230° + cos230° is:
(A) 0
(B) 2
(C) 3
(D) 1
Solution:
Correct answer: (D)
sin230° + cos230°
Using the identity sin2A + cos2A = 1,
= 1
Question 4: The value of cosec255° – cot255° is:
(A) 1
(B) 2
(C) 3
(D) 0
Solution:
Correct answer: (A)
cosec255° – cot255°
Using the identity, cot2A + 1 = cosec2A (or cosec2A – cot2A = 1),
= 1
Question 5: If cot A = 20/21, then the value of cosec A is:
(A) 21/20
(B) 20/29
(C) 29/21
(D) 21/29
Solution:
Correct answer: (C)
Given,
cot A = 20/21
We know that, cosec2A = cot2A + 1
cosec2A = (20/21)2 + 1
= (400/441) + 1
= (400 + 441)/441
= 841/441
cosec A = √(841/441) = 29/21
Question 6: The value of (sec 40° + tan 40°)(sec 40° – tan 40°) is:
(A) -1
(B) 1
(C) cos 40°
(D) sin 40°
Solution:
Correct answer: (B)
(sec 40° + tan 40°)(sec 40° – tan 40°) = sec240° – tan240°
Using the identity 1 + tan2A = sec2A,
= 1
Question 7: The value of 1/(sin θ – tan θ) is:
(A) cot θ/(cos θ – 1)
(B) cot θ/(cot θ – cosec θ)
(C) cosec θ – cot θ
(D) cot θ
Solution:
Correct answer: (A)
1/(sin θ – tan θ)
= 1/ [sin θ – (1/ cot θ)]
= cot θ/(sin θ cot θ – 1)
= cot θ/ [sin θ × (cos θ/ sin θ) – 1]
= cot θ/(cos θ – 1)
Question 8: The value of (sec A – 1)/ (sec A + 1) is:
(A) (1 + cos A)/ (1 – cos A)
(B) (cos A – 1)/ (1 + cos A)
(C) (1 – cos A)/ (1 + cos A)
(D) (cos A – 1)/ (1 – cos A)
Solution:
Correct answer: (C)
(sec A – 1)/ (sec A + 1)
= [(1/cos A) – 1]/ [(1/cos A) + 1]
= [(1 – cos A)/cos A]/ [(1 + cos A)/cos A]
= [(1 – cos A) / cos A] × [cos A / (1 + cos A)]
= (1 – cos A)/(1 + cos A)
Question 9: The value of cot2θ – (1/sin2θ) is:
(A) 2
(B) 1
(C) 0
(D) -1
Solution:
Correct answer: (D)
cot2θ – (1/sin2θ)
= cot2θ – cosec2θ
= -1(cosec2θ – cot2θ)
Using the identity cot2A + 1 = cosec2A,
= -1 (1)
= -1
Question 10: f cos θ = 3/5, then evaluate (sin θ – cot θ)/2 tan θ.
Solution:
Given,
cos θ = 3/5
Using the identity sin2θ + cos2θ = 1,
sin2θ = 1 – cos2θ
= 1 – (3/5)2
= 1 – (9/25)
= (25 – 9)/25
= 16/25
sin θ = √(16/25) = 4/5
tan θ = sin θ/cos θ
= (4/5) / (3/5)
= (4/5) (5/3)
= 4/3
cot θ = 1/tan θ = 3/4
(sin θ – cot θ)/2 tan θ = [(4/5) – (3/4)]/ [2 × (4/3)]
= [(16 – 15)/20] / (8/3)
= (1/20) × (3/8)
= 3/160
Question 11: If cot θ = b/a, then evaluate (cos θ + sin θ)/ (cos θ – sin θ).
Solution:
Given,
cot θ = b/a
Now,
(cos θ + sin θ)/ (cos θ – sin θ)
Dividing the numerator and denominator by sin θ,
= [(cos θ/sin θ) + (sin θ/sin θ)] / [(cos θ/sin θ) – (sin θ/sin θ)]
= (cot θ + 1)/ (cot θ – 1)
Substituting cot θ = b/a,
= [(b/a) + 1]/ [(b/a) – 1]
= [(b + a)/a] / [(b – a)/a]
= (b + a)/(b – a)
Question 12: Prove that: √(sec2A + cosec2A) = tan A + cot A
Solution:
LHS = √(sec2A + cosec2A)
Using the identities: 1 + tan2A = sec2A and cot2A + 1 = cosec2A,
= √(1 + tan2A + 1 + cot2A)
= √(tan2A + cot2A + 2 × 1)
= √(tan2A + cot2A + 2 tan A cot A) (since tan A × cot A = 1)
= √(tan A + cot A)2
= tan A + cot A
= RHS
Thus, √(sec2A + cosec2A) = tan A + cot A
Hence proved.
Question 13: Prove that: (1 + sec θ)/sec θ = sin2θ/(1 – cos θ)
Solution:
LHS = (1 + sec θ)/sec θ
= (1/sec θ) + (sec θ/sec θ)
= cos θ + 1 ….(i)
RHS = sin2θ/(1 – cos θ)
By rationalising the denominator,
= [sin2θ/(1 – cos θ)] × [(1 + cos θ)/(1 + cos θ)]
= sin2θ(1 + cos θ) / (1 – cos2θ)
Using the identity sin2θ + cos2θ = 1,
= sin2θ(1 + cos θ)/sin2θ
= 1 + cos θ ….(ii)
From (i) and (ii),
(1 + sec θ)/sec θ = sin2θ/(1 – cos θ)
Hence proved.
Question 14: Prove that: (tan A + sec A – 1)/ (tan A – sec A + 1) = tan A + sec A
Solution:
LHS = (tan A + sec A – 1)/ (tan A – sec A + 1)
Using the identity 1 + tan2A = sec2A,
= [(tan A + sec A) – (sec2A – tan2A)] / (tan A – sec A + 1)
= [(tan A + sec A) – (sec A – tan A)(sec A + tan A)] / (tan A – sec A + 1)
= [(tan A + sec A)(1 – sec A + tan A)]/ (tan A – sec A + 1)
= tan A + sec A
= RHS
Hence proved.
Question 15: Prove that:
(i) tan2θ – cot2θ = sec2θ – cosec2θ
(ii) cos4θ – sin4θ = 1 – 2 sin2θ
Solution:
(i) tan2θ – cot2θ = sec2θ – cosec2θ
LHS = tan2θ – cot2θ
Using the identities: 1 + tan2A = sec2A, 1 + cot2A = cosec2A,
= (sec2θ – 1) – (cosec2θ – 1)
= sec2θ – 1 – cosec2θ + 1
= sec2θ – cosec2θ
= RHS
Hence proved.
(ii) cos4θ – sin4θ = 1 – 2 sin2θ
LHS = cos4θ – sin4θ
= (cos2θ)2 – (sin2θ)2
= (cos2θ + sin2θ)(cos2θ – sin2θ)
Using the identity sin2θ + cos2θ = 1,
= (1)[(1 – sin2θ) – sin2θ]
= 1 – 2 sin2θ
= RHS
Hence proved.
Question 16: Prove that: √[(cosec2θ – 1)/ cosec2θ] = cos θ
Solution:
LHS = √[(cosec2θ – 1)/ cosec2θ]
Using the identity 1 + cot2θ = cosec2θ,
= √(cot2θ/ cosec2θ)
= cot θ/cosec θ
= (cos θ/sin θ) × sin θ
= cos θ
= RHS
Hence proved.
Question 17: If sin A = 3/5, then find the value of cos A and cot A.
Solution:
Given,
sin A = 3/5
We know that,
sin2A + cos2A = 1
cos2A = 1 – sin2A
= 1 – (3/5)2
= 1- (9/25)
= (25 – 9)/25
= 16/25
cos A = √(16/25) = 4/5
cot A = cos A/sin A
= (4/5)/ (3/5)
= 4/3
Therefore, cos A = 4/5 and cot A = 4/3
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