Trigonometry
- ABC is a right angled triangle, right angled at B and ∠A = 60° and AB = 20 cm, then the ratio of sides BC and CA is
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∠B = 90°
∠A = 60°
∠C = 180° – 90° – 60° = 30°cosC = BC CA ⇒cos30° = BC CA ⇒ √3 = BC = √3 : 2 2 CA
Correct Option: D
∠B = 90°
∠A = 60°
∠C = 180° – 90° – 60° = 30°cosC = BC CA ⇒cos30° = BC CA ⇒ √3 = BC = √3 : 2 2 CA
- If tan 2θ . tan 3θ = 1, where 0° < θ < 90° then the value of θ is
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tan2θ . tan3θ = 1
⇒ tan3θ = 1 = cot2θ tan2θ
⇒ tan3θ = tan (90° – 2θ)
⇒ 3θ = 90° – 2θ
⇒ 3θ + 2θ = 5θ = 90°⇒ θ = 90° = 18° 5
Correct Option: B
tan2θ . tan3θ = 1
⇒ tan3θ = 1 = cot2θ tan2θ
⇒ tan3θ = tan (90° – 2θ)
⇒ 3θ = 90° – 2θ
⇒ 3θ + 2θ = 5θ = 90°⇒ θ = 90° = 18° 5
- If cos²α – sin²α = tan²β, then the value of cos²β – sin²β is
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cos²α – sin²α = tan²β
⇒ cos²α – (1 – cos²α) = tan²β
⇒ 2cos²α – 1 = tan²β
⇒ 2cos²α = 1 + tan²β = sec²β⇒ cos²β = 1 2cos²α
sin²β = 1 – cos²β= 1 - 1 2cos²α = 2cos²α - 1 2cos²α
∴ cos²β – sin²β= 1 - 2cos²α - 1 2cos²α 2cos²α = 1 - 2cos²α + 1 2cos²α = 2(1 - cos²α) = sin²α 2cos²α cos²α
= tan²α
Note : It is an identity.Correct Option: C
cos²α – sin²α = tan²β
⇒ cos²α – (1 – cos²α) = tan²β
⇒ 2cos²α – 1 = tan²β
⇒ 2cos²α = 1 + tan²β = sec²β⇒ cos²β = 1 2cos²α
sin²β = 1 – cos²β= 1 - 1 2cos²α = 2cos²α - 1 2cos²α
∴ cos²β – sin²β= 1 - 2cos²α - 1 2cos²α 2cos²α = 1 - 2cos²α + 1 2cos²α = 2(1 - cos²α) = sin²α 2cos²α cos²α
= tan²α
Note : It is an identity.
- If tan (A + B) = √AAAA and tan (A – B) = (1 / √3) , ∠A + ∠B) < 90°, ∠A ≥ B, then ∠A is
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tan(A + B) = √3 = tan60°
⇒ A + B = 60° ...(i)tan(A – B) = 1 = tan30° √30
⇒ A – B = 30° ...(ii)
∴ A + B + A – B = 60° + 30°
⇒ 2A = 90°⇒ A = 90° = 45° 2
Correct Option: C
tan(A + B) = √3 = tan60°
⇒ A + B = 60° ...(i)tan(A – B) = 1 = tan30° √30
⇒ A – B = 30° ...(ii)
∴ A + B + A – B = 60° + 30°
⇒ 2A = 90°⇒ A = 90° = 45° 2
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The value of sin θ - 2sin3 θ is 2 cos3 θ - cosθ
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Expression
= sinθ - 2sin3θ 2cos3θ -cosθ = sinθ ( 1 - 2sin2θ) cosθ (2cos2θ - 1) = sinθ . (1 - 2 (1 - cos²θ))) cos θ (2cos²θ - 1) = tan θ (1 + 2 cos² θ - 2) (2 cos² θ - 1) = tan θ. 2 cos²θ - 1 = tan θ 2 cos²θ - 1
Correct Option: C
Expression
= sinθ - 2sin3θ 2cos3θ -cosθ = sinθ ( 1 - 2sin2θ) cosθ (2cos2θ - 1) = sinθ . (1 - 2 (1 - cos²θ))) cos θ (2cos²θ - 1) = tan θ (1 + 2 cos² θ - 2) (2 cos² θ - 1) = tan θ. 2 cos²θ - 1 = tan θ 2 cos²θ - 1
