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If θ is a positive acute angle and 4 cos2 θ – 4 cos θ + 1 = 0, then the value of tan (θ – 15°) is equal to (SSC CHSL 2014)
- a)0
- b)1
- c)√3
- d)1/√3
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
If θ is a positive acute angle and 4 cos2 θ – 4 cos ...
Given Equation:
4 cos² θ – 4 cos θ + 1 = 0
Key Points:
- The given equation represents a quadratic equation in terms of cos θ.
- Let's denote cos θ as x for simplicity.
Solution:
Step 1: Solve the Quadratic Equation
- Substitute x for cos θ in the given equation:
4x² – 4x + 1 = 0
- This equation can be factored as:
(2x – 1)² = 0
- Solving for x, we get:
x = 1/2
Step 2: Find cos(θ – 15°)
- Using the compound angle formula for cosine, we have:
cos(θ – 15°) = cos θ cos 15° + sin θ sin 15°
- Substituting the values of cos θ (x) and sin θ (√(1 – x²)), we get:
cos(θ – 15°) = (1/2)(√3/2) + (√3/2)(1/2)
cos(θ – 15°) = √3/4 + √3/4
cos(θ – 15°) = √3/2
Step 3: Find tan(θ – 15°)
- Using the trigonometric identity tan θ = sin θ / cos θ, we have:
tan(θ – 15°) = sin(θ – 15°) / cos(θ – 15°)
- Substituting sin(θ – 15°) = sin θ cos 15° – cos θ sin 15° and cos(θ – 15°) = √3/2, we get:
tan(θ – 15°) = (√3/2 – √3/2) / (√3/2)
tan(θ – 15°) = 0 / (√3/2)
tan(θ – 15°) = 0
Therefore, the value of tan(θ – 15°) is 0. Hence, the correct answer is option 'a) 0'.
4 cos² θ – 4 cos θ + 1 = 0
Key Points:
- The given equation represents a quadratic equation in terms of cos θ.
- Let's denote cos θ as x for simplicity.
Solution:
Step 1: Solve the Quadratic Equation
- Substitute x for cos θ in the given equation:
4x² – 4x + 1 = 0
- This equation can be factored as:
(2x – 1)² = 0
- Solving for x, we get:
x = 1/2
Step 2: Find cos(θ – 15°)
- Using the compound angle formula for cosine, we have:
cos(θ – 15°) = cos θ cos 15° + sin θ sin 15°
- Substituting the values of cos θ (x) and sin θ (√(1 – x²)), we get:
cos(θ – 15°) = (1/2)(√3/2) + (√3/2)(1/2)
cos(θ – 15°) = √3/4 + √3/4
cos(θ – 15°) = √3/2
Step 3: Find tan(θ – 15°)
- Using the trigonometric identity tan θ = sin θ / cos θ, we have:
tan(θ – 15°) = sin(θ – 15°) / cos(θ – 15°)
- Substituting sin(θ – 15°) = sin θ cos 15° – cos θ sin 15° and cos(θ – 15°) = √3/2, we get:
tan(θ – 15°) = (√3/2 – √3/2) / (√3/2)
tan(θ – 15°) = 0 / (√3/2)
tan(θ – 15°) = 0
Therefore, the value of tan(θ – 15°) is 0. Hence, the correct answer is option 'a) 0'.
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Community Answer
If θ is a positive acute angle and 4 cos2 θ – 4 cos ...
4 cos2θ – 4 cosθ + 1 = 0
(2 cosθ – 1)2 = 0
or, 2 cosθ = 1
⇒ cosθ = 1/2 ; θ = 60°
Hence, the value of tan (θ – 15°) = tan (60° – 15°)
= tan 45° = 1
(2 cosθ – 1)2 = 0
or, 2 cosθ = 1
⇒ cosθ = 1/2 ; θ = 60°
Hence, the value of tan (θ – 15°) = tan (60° – 15°)
= tan 45° = 1
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If θ is a positive acute angle and 4 cos2 θ – 4 cos θ + 1 = 0, then the value of tan (θ – 15°) is equal to (SSC CHSL 2014)a)0b)1c)√3d)1/√3Correct answer is option 'B'. Can you explain this answer?
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