Maths WBBSE Class 10 Solutions Chapter 23 Trigonometric Ratios And Trigonometric Identities Exercise 23.3
Question 1. If sinθ = then let us write the value of cosec/1+cotθ by determining it.
Solution : sinθ =4/5
∴ cosecθ =1/sinθ
=5/4
cot2θ = cosec2θ – 1 = (5/4)²-1
= 25/16-1
= 25-16 / 16
= 9/16
∴ cosθ= √9/16
= 3/4
Now, cosecθ / 1+cotθ = 5/4 / 1+3/4
= 5/4 x 4/7
= 5/7
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Question 2. If tanθ = 3/4, then let us show that √(1-sinθ/1+sinθ) = 1/2
Solution: tanθ = 3/4
let AB =3K
& BC = 4K
∴ AC² = AB²+BC²
= (3K)²+(4K)² = 9K²+16K²
= 25K²
.. AC = √25K²
= 5K
∴ sinθ = AB/AC = 3K/5K = 3/5
= 1/2 = R.H.S Proved
Maths WBBSE Class 10 Solutions
Question 3. If tanθ = 1, then let us determine the value of 8sinθ+5cosθ/sin³ θ-2cos³θ+7 cosθ
Solution: tanθ = 1
Question 2.
1. Let us express cosecθ and tanθ in terms of sine.
Solution: cosecθ= 1/sinθ
tanθ= sin/cos
= sinθ/√1-sin²θ
2. Let us write cosecθ and tanθ in terms of cosθ
Solution: cosecθ =1/sinθ = 1/√1-cos²θ
tanθ = sinθ/cosθ
= √1-cos²θ/cosθ
Maths WBBSE Class 10 Solutions
Question 3.
1. If secθ + tanθ = 2, then let us determine the value of (secθ– tanθ).
Solution: We know, sec²θ tan²θ = 1
or, (secθ+tanθ) (secθ-tanθ) = 1
or, 2 x (secθ – tanθ) = 1
secθ – tanθ = 1/2
2. If cosecê – cote = √2-1, then let us write by calculating, the value of (cosecθ + cotθ).
Solution: We know, cosec²θ – cot²θ
or, (cosecθ+cotθ) (cosecθ- cotθ) = 1
or, (cosecθ+cotθ) (√2+1) = 1
∴ cosecθ+cotθ = 1/√2-1
= (√2+1)/(√2-1)(√2+1)
= (√2+1)/2-1 = √2+1
3. If sinθ+cosθ = 1, then let us determine the value of sine x cosθ.
Solution: sinθ + cosθ = 1
or, (sinθ+cosθ)² = 12
or, sin²θ+ cos²θ + 2sinθ x cosθ = 1
or, 1+2sinθ cosθ = 1 [as sin²θ + cos²θ = 1]
∴ 2sinθ cosθ = 0 => sinθ cosθ = 0.
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4. If tanθ+cotθ = 1, then let us determine the value of (tanθ – cotθ).
Solution: (tanθ-cotθ)²= (tanθ + cotθ) – 4. tanθ . cotθ
=(2)² – 4 x tanθ x 1/tan θ
=4-4=0.
∴tanθ – cotθ = 0
5. If sinθ – cosθ = 7/13 then let us determine the value of sinθ + cosθ.
Solution: (sinθ cosθ)² = (7/13)²
or, sin²θ+ cos²θ – 2sinθ cosθ = 49/169
or, 1 – 2sinθ cosθ = 49/169
or, 1-169 = 2sinθ cosθ
or, 169-49/169 = 2sinθ cosθ
or, 120/169 = 2sinθ cosθ
∴ (sine+cose)²= sin²θ+ cos²θ + 2sinθ cosθ
= 1 + 120/169
= 120+169/169
= 289/169
∴ sinθ+cosθ = √289/169
=17/13
6. If sinθcosθ = 1/2, then let us write by calculating, the value of (sinθ + cosθ).
Solution: (sinθ+cosθ)²= sin²θ+ cosθ²+2. sinθ. cosθ
= 1+2×1/2 =2.
∴ sinθ cosθ = √2.
West Bengal Board Class 10 Math Book Solution In English
7. If secθ – tanθ, then let us determine the values of both secθ and tanθ.
Solution: We know, sec²θ-tan²θ = 1
8. If cosecθ+cotθ = √3, then let us determine the values of both cosec e and cote.
Solution: We know, cosec²- cot² = 1
9. If sin θ + cos θ/ sin θ-cos θ =7, then let us write by calculating, the value of tanθ.
Solution: sin θ + cos θ / sinθ-cos θ =7
WBBSE Solutions Guide Class 10
10. If cosec θ + sin θ/cosecθ-sin θ = 5/3, then let us write by calculating, the value of sine.
Solution: cosecθ+ sinθ/cosecθ -sinθ= 5/3
11. If secθ+cosθ = 5/3 then let us write by calculating, the value of (secθ – cosθ).
Solution: (secθ – cosθ)²= (secθ+cosθ)² – 4. secθ. cosθ
= (5/3)²- 4 . sec x 1/secθ
= 25/9 – 4
= 25- 1 / 9
12. Let us determine the value of tan from the relation 5sin²0 + 4cos²0 = 9/2
Solution: 5sin²θ + 4cos²θ = 9/2
WBBSE Solutions Guide Class 10
13. If tan²θ+ cot²θ = 10/3, then let us determine the values of tan e + cote and tan e – cote and from these let us write the value of tanθ.
Solution: (tanθ+cotθ)² = tan²θ+ cot²θ + 2tanθ. cotθ
14. If sec²θ + tan²θ =13/12 then let us write by calculating, the value of (sec10 – tan10).
Solution: sec4θ– tan4θ
= (secθ+tan²θ) (sec²θ – tan²θ)
= 13/12 x 1
= 13/12
WBBSE Solutions Guide Class 10
Question 4.
1. In ΔPQR, ∠Q is a right angle. If PR = √5 units and PQ – RQ = 1 unit, then let us determine the value of cos P – cos R.
Solution: cos P – cos R (given PQ – QR = 1)
= PQ/PR – QR/PR
= PQ-QR / PR
= 1√5 unit
2. In ΔXYZ, Y is a right angle. If XY = 2√3 units and XZ – YZ = 2 units then let us determine the value of (sec X – tan X).
Solution: Given XY = 2√3 unit
& XZ – YZ = 2 units.
Question 5. Let us eliminate ‘θ’ from the relations’
1. x = 2 sinθ, y = 3 cosθ
Solution: x = 2sinθ & y = 3coseθ
∴ sinθ = x/2 ∴ cosθ = y/3
We know, sin²θ + cos²θ = 1
∴ x²/4 + y2/9 = 1
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2. 5x=3secθ, y = 3tanθ
Solution : 5x = 3 secθ; &y = 3tanθ
Question 6.
1. If sinα = 5/13 then let us show that, tanα+ secα = 1.5.
Solution: sinα = 5/13
2. If tanA = n/m then let us determine the values of both sinA and secA.
Solution: tanA = n/m
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3. If sec = x/√x² + y² then let us show that, x sinθ = y cosθ.
Solution:
4. If sinα = a²-b² / a²+b² then let us show that, cotα = 2ab/a²-b²
Solution: sinα = a²-b² / a²+b²
5. If sinθ / x = cosθ / y then let us show that sinθ = x-y/√x² + y²·
Solution: sinθ / x = cos θ/ y = K(let)
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6. If (1 + 4x²) cosA = 4x, then let us show that cosecA + cotA = 1+2x/1-2x
Solution: (1+ 4x²)cosA = 4x
7. If x = a sine and y = b tanθ, then let us prove that a²/x²- b²/y² = 1
Solution: x = a sine & y = b tane
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8. If sinθ + sin² = 1, then let us prove that cos²θ+ cos²θ = 1.
Solution: sinθ + sin²θ = 1
∴ sinθ = 1 – sin²θ = cos²θ
∴ sin20 = cos10
∴ cos20+ cos 0 sine + sin20 = 1 Proved
West Bengal Board Class 10 Math Book Solution In English Chapter 23 Trigonometric Ratios And Trigonometric Identities Exercise 23.3 Multiple Choice Question
1. If 3x = cosecα and 3/x = cot α ,then the value of 3(x² – 1/x²) is
1. 1/27
2. 1/81
3. 1/3
4. 1/9
Solution: cosecα = 3x & cot = 3/x
∴ 3x² – 3/x²
= 1/3
Answer: 3. 1/3
2. If 2x = sec A and 2/x = tan A, then the value of 2(x² – 1/x²)is
1. 1/2
2. 1/4
3. 1/8
4. 1/16
Solution: sec²A – tan²A= 1
Answer: 1. 1/2
3. If tan∝ + cot ∝ = 2, then the value of (tan¹³∝ + cot¹³∝) is
1. 1
2. 0
3. 2
4. None
Solution: tan∝ + cot∝ = 2
Answer: 3. 2
4. If sin 0 – cos 0 = 0 (0º ≤θ≤ 90º) and sec θ + cosec θ = x, then the value of x is
1. 1
2. 2
3. √2
4. 2√2
Solution: sinθ- cosθ = 0
Answer: 4. 2√2
5. If 2cos³θ = 1, then the value of θ is
1. 10º
2. 15º
3. 20º
4. 30º
Solution: 2cos³θ = 1
∴ cos3θ = 1/2 = cos60º
∴ 3θ = 60º .. θ= 60º /2 = 20º
Answer: 3. 20º
West Bengal Board Class 10 Math Book Solution In English Chapter 23 Trigonometric Ratios And Trigonometric Identities Exercise 23.3 True Or False
1. If 0°< a < 90°, then the least value of (sec²α + cos²α) is 2.
Answer: True
2. The value of (cos0° x cos1° x cos2° x cos3° x…………..cos90°) is 1.
Answer: False
Class 10 WBBSE Math Solution In English Chapter 23 Trigonometric Ratios And Trigonometric Identities Exercise 23.3 Fill In The Blanks
1. The value of (4/sec²θ+ 1/1+ cot²θ +3sin²θ) is————-
Solution: 4/sec² 0 + 1/1+ cot²0 +3sin²0
= 4cos²0 + sin²0 + 3sin²0
= 4(cos²0 + sin²0)
= 4 x 1
= 4
2. If sin (θ-30°) = 1/2 then the value of cos e is————
Solution: sin(θ – 30°) = 1/2 = sin30°
θ – 30° = 30°
θ = 30° + 30° ∴ θ =- 60°
∴ cos θ= cos 60° = 1/2
3. If cos²θ – sin²θ = 1/2 , then the value of Cos4θ- sin4θ is ———
Solution: Cos4θ- sin4θ
(cos²θ)² – (sin²θ)²
= (cos²θ + sin²θ)(cos²θ – sin²θ)
= 1 x 1/2
= 1/2
Chapter 23 Trigonometric Ratios And Trigonometric Identities Exercise 23.3 Short Answers
1. If r cosθ = 2√3, r sinθ = 2, and 0º <θ <90º, then let us determine the values of both r and θ
Solution: r cosθ = 2√3, r sinθ = 2
2. If sinA + sinB = 2 where 0° SA≤ 90° and 0° B≤ 90°, then let us find out the value of (cosA+ cosB).
Solution. sinA + sinB = 2
Greatest value of sinA ∴ sin A = sin90° ∴ A= 90°
Greatest value of sinB ∴ sin B = sin90° ∴ B = 90°
cosA + cosB = cos90° + cos90° = 0 + 0 = 0
3. If 0°<θ< 90°, then let us calculate the least value of (9tan²θ + 4cot2θ).
Solution. 9tan²θ+4cot²θ = (3tanθ)² + (2cot²θ)²
= (3tanθ2-cotθ)²-2. 3tanθ. 2cotθ
= (3tanθ-2cotθ)²-12 tanθ x 1/tanθ
∴ Least value of (9tan²θ+ 4cot²θ) is 12.
4. Let us calculate the value of ( Sin6+cos6+ 3sin²α cos²α).
Solution: Sin6∝+cos6∝+3sin²α cos²α.
= (sin²α)³ + (cos²α)³ + 3. sin²a. cos²α(sin²α + cos²α)
= (sin²α + cos²α)³ = (1)³
= 1.
5. If cosec20 = 2cote and 0° <0< 90°, then let us determine the value of 0.
Solution. cosec²0 = 2cote
or, 1+ cot²θ – 2cotθ = 0
or, (1 – cote)² = 0
1-cotθ=0 .. cotθ1 = cot45°
∴ 0 = 45°