Trigonometry
- 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 (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 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
- 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 tan A = n tan B and sin A = m sin B, then the value of cos²A is
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tan A = n tan B
⇒ tan B = 1 tan A n ⇒ cot B = n tan A
and, sin A = m sin B⇒ sin B = 1 sinA m ⇒ cosec B = 1 m sin A
∵ cosec²B – cot²B = 1⇒ m² - n² = 1 sin²A tan²A ⇒ m² - n²cos²A = 1 sin²A sin²A ⇒ m² - n²cos²A = 1 sin²A
⇒ m² – n² cos²A = sin²A
⇒ m² – n² cos²A = 1 – cos²A
⇒ m² –1 = n² cos²A – cos²A
⇒ m² – 1 = (n² – 1) cos²A⇒ cos²A = m² - 1 n² - 1
Correct Option: D
tan A = n tan B
⇒ tan B = 1 tan A n ⇒ cot B = n tan A
and, sin A = m sin B⇒ sin B = 1 sinA m ⇒ cosec B = 1 m sin A
∵ cosec²B – cot²B = 1⇒ m² - n² = 1 sin²A tan²A ⇒ m² - n²cos²A = 1 sin²A sin²A ⇒ m² - n²cos²A = 1 sin²A
⇒ m² – n² cos²A = sin²A
⇒ m² – n² cos²A = 1 – cos²A
⇒ m² –1 = n² cos²A – cos²A
⇒ m² – 1 = (n² – 1) cos²A⇒ cos²A = m² - 1 n² - 1
