Trigonometry
- a, b, c are the lengths of three sides of a triangle ABC. If a, b, c are related by the relation a² + b² + c² = ab + bc + ca, then the value of sin²A + sin²B + sin²C is
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a² + b² + c² = ab + bc + ca
⇒ 2a² + 2b² + 2c² = 2ab + 2bc + 2ca
⇒ a² + b² + b² + c² + c² + a² – 2ab – 2bc – 2ca = 0
⇒ a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca = 0
⇒ (a – b)² + (b – c)² + (c – a)² = 0
⇒ a – b = 0
⇒ a = b
b – c = 0
⇒ b = c
c – a = 0
⇒ c = a
∴ ∆ ABC is an equilateral triangle.
∴ ∠A = ∠B = ∠ C = 60°
∴ sin²A + sin²B + sin²C = 3 sin²A = 3 × sin² 60°=3 × √3 ² 2 = 3 × 3 = 9 4 4
Correct Option: D
a² + b² + c² = ab + bc + ca
⇒ 2a² + 2b² + 2c² = 2ab + 2bc + 2ca
⇒ a² + b² + b² + c² + c² + a² – 2ab – 2bc – 2ca = 0
⇒ a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca = 0
⇒ (a – b)² + (b – c)² + (c – a)² = 0
⇒ a – b = 0
⇒ a = b
b – c = 0
⇒ b = c
c – a = 0
⇒ c = a
∴ ∆ ABC is an equilateral triangle.
∴ ∠A = ∠B = ∠ C = 60°
∴ sin²A + sin²B + sin²C = 3 sin²A = 3 × sin² 60°=3 × √3 ² 2 = 3 × 3 = 9 4 4
- If x = a secθ, y = b tanθ, then
x² - y² is a² b²
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x = a sec θ ⇒ = x = sec θ a and y = b sec θ ⇒ = y = tan θ b ∴ x² - y² = sec² θ – tan² θ = 1 a² b²
Correct Option: C
x = a sec θ ⇒ = x = sec θ a and y = b sec θ ⇒ = y = tan θ b ∴ x² - y² = sec² θ – tan² θ = 1 a² b²
- If sin (60° – x) = cos (y + 60°), then the value of sin (x – y) is
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sin (60° – x) = cos (y + 60°)
⇒ sin (60°–x) = sin (90°– y – 60°) [ ∵ sin (90° – θ) = cos θ)
⇒ 60° – x = 90° – y – 60° = 30° – y
⇒ x – y = 60° – 30°
⇒ x – y = 30°∴ sin (x - y) = sin 30° = 1 2
Correct Option: B
sin (60° – x) = cos (y + 60°)
⇒ sin (60°–x) = sin (90°– y – 60°) [ ∵ sin (90° – θ) = cos θ)
⇒ 60° – x = 90° – y – 60° = 30° – y
⇒ x – y = 60° – 30°
⇒ x – y = 30°∴ sin (x - y) = sin 30° = 1 2
- If sin (3α – β) = 1 and cos (2α+β) = (1 / 2) , then the value of tan α is
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sin (3 α – β) = 1 = sin 90°
⇒ 3α – β = 90° ... (i)cos (2α + β) = 1 = cos 60° 2
⇒ 2 α + β = 60° .... (ii)
By adding both equations,
3α + 2α = 90° + 60°
⇒ 5α = 150⇒ α = 150 = 30° 5 ∴ tan α = tan 30° = 1 √3
Correct Option: B
sin (3 α – β) = 1 = sin 90°
⇒ 3α – β = 90° ... (i)cos (2α + β) = 1 = cos 60° 2
⇒ 2 α + β = 60° .... (ii)
By adding both equations,
3α + 2α = 90° + 60°
⇒ 5α = 150⇒ α = 150 = 30° 5 ∴ tan α = tan 30° = 1 √3
- If sin θ + cos θ = p and sec θ + cosec θ = θ, then the value of θ (p² – 1) is
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sin θ + cos θ = p
sec θ + cosec θ = q⇒ 1 + 1 = q cos θ sinθ ⇒ sinθ + cosθ = q sinθ . cosθ ∴ q(p² - 1) = sin θ + cos θ ((sinθ + cosθ)² - 1) sinθ . cosθ = sinθ + cosθ . (sin²θ + cos²θ + 2sinθ.cosθ - 1) sinθ . cosθ = sinθ + cosθ . 2sinθ . cosθ sinθ . cosθ
= 2pCorrect Option: C
sin θ + cos θ = p
sec θ + cosec θ = q⇒ 1 + 1 = q cos θ sinθ ⇒ sinθ + cosθ = q sinθ . cosθ ∴ q(p² - 1) = sin θ + cos θ ((sinθ + cosθ)² - 1) sinθ . cosθ = sinθ + cosθ . (sin²θ + cos²θ + 2sinθ.cosθ - 1) sinθ . cosθ = sinθ + cosθ . 2sinθ . cosθ sinθ . cosθ
= 2p