Trigonometry
- If α + β = 90°, then the expression (tan α / tan β) + sin²α + sin²β is equal to :
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α + β = 90°
⇒ α = 90° – β
⇒ tanα = tan (90° – β) = cot β.
sinα = sin (90° – β) = cosβ
∴ Expression= cot β + tan² β + sin² β tan β
= cot² β + 1
= cosec² β = cosec² (90° – α)
= sec²αCorrect Option: D
α + β = 90°
⇒ α = 90° – β
⇒ tanα = tan (90° – β) = cot β.
sinα = sin (90° – β) = cosβ
∴ Expression= cot β + tan² β + sin² β tan β
= cot² β + 1
= cosec² β = cosec² (90° – α)
= sec²α
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Value of the expression : 1 + 2sin 60° cos60° + 1 - 2sin 60° cos60° is sin 60° + cos60° sin 60° - cos60°
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Expression
2 + √3 + 2 - √3 √3
+ 1√3
- 1= (2 + √3)(√3 - 1) + (√3 + 1)(2 - √3) (√3 + 1)(√3 - 1) = 2√3 - + 3 - √3 + 2√3 - + 2 - √3 3 - 1 = 4√3
- 2√3= 2√3 = √3 2 2
Correct Option: C
Expression
2 + √3 + 2 - √3 √3
+ 1√3
- 1= (2 + √3)(√3 - 1) + (√3 + 1)(2 - √3) (√3 + 1)(√3 - 1) = 2√3 - + 3 - √3 + 2√3 - + 2 - √3 3 - 1 = 4√3
- 2√3= 2√3 = √3 2 2
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If sin 2θ - √3 then the value of sin3θ is equal to (Take 0° ≤ θ ≤ 90° ) 2
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sin2θ = √3 = sin 60° 2
⇒ 2θ = 60°
⇒ θ = 30°
∴ sin 3θ = sin 90° = 1Correct Option: B
sin2θ = √3 = sin 60° 2
⇒ 2θ = 60°
⇒ θ = 30°
∴ sin 3θ = sin 90° = 1
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If sin θ + cosθ = 3 then the value of sin4 q is : sin θ - cosθ
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sinθ + cosθ = 3 sinθ - cosθ 1
By componendo and dividendosin θ + cos θ + sin θ – cosθ = 3 + 1 sin θ + cos θ - sin θ + cosθ 3 - 1 ⇒ 2 sinθ = 4 2 cosθ 2
⇒ tanθ = 2∴ cotθ = 1 2
∴ cosecθ = √1 + cot² θ
∴ sin θ = 2 √5 sin4θ = 16 25
Correct Option: E
sinθ + cosθ = 3 sinθ - cosθ 1
By componendo and dividendosin θ + cos θ + sin θ – cosθ = 3 + 1 sin θ + cos θ - sin θ + cosθ 3 - 1 ⇒ 2 sinθ = 4 2 cosθ 2
⇒ tanθ = 2∴ cotθ = 1 2
∴ cosecθ = √1 + cot² θ∴ sin θ = 2 √5 sin4θ = 16 25
- If cos A + sin A = √2 cos A then cos A – sin A is equal to : (where 0° < A < 90°)
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cosA + sinA = 2 cosA --- (i)
cosA – sinA = x (let) --- (ii)
On squaring both equation and adding
cos²A + sin²A + 2 sinA . cosA + cos²A + sin²A – 2 sinA cosA = 2
cos²A + x ²
⇒ 2 (cos²A + sin²A) = 2 cos²A + x 2
⇒ x² + 2 cos²A = 2
⇒ x² = 2 – 2 cos²A
= 2 (1 – cos²A) = 2 sin²A
∴ x = √2 sin ACorrect Option: A
cosA + sinA = 2 cosA --- (i)
cosA – sinA = x (let) --- (ii)
On squaring both equation and adding
cos²A + sin²A + 2 sinA . cosA + cos²A + sin²A – 2 sinA cosA = 2
cos²A + x ²
⇒ 2 (cos²A + sin²A) = 2 cos²A + x 2
⇒ x² + 2 cos²A = 2
⇒ x² = 2 – 2 cos²A
= 2 (1 – cos²A) = 2 sin²A
∴ x = √2 sin A
