Trigonometry
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If cos α = n and cos α = m , then the vlaue of cos² β is sin β cos β
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cos α = n and cos α = m sin β sin β
⇒ cosα = n sinβ and cosα = m cosβ.
∴ n² sin² β = m² cos² β
⇒ n² (1 – cos² β) = m² cos² β
⇒ n² – n² cos² β = m² cos² β ⇒ m² cos² β + n² cos² β = n²
⇒ cos² β (m² + n² ) = n²⇒ cos²β = n² m² + n²
Correct Option: C
cos α = n and cos α = m sin β sin β
⇒ cosα = n sinβ and cosα = m cosβ.
∴ n² sin² β = m² cos² β
⇒ n² (1 – cos² β) = m² cos² β
⇒ n² – n² cos² β = m² cos² β ⇒ m² cos² β + n² cos² β = n²
⇒ cos² β (m² + n² ) = n²⇒ cos²β = n² m² + n²
- The value of tan1°tan2°tan3° ........tan89° is
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tan (90° – θ) = cotθ tanθ.cotθ = 1
tan 89° = tan (90° – 1°) = cot 1°.
tan 88° = tan (90° – 2°) = cot 2°.
∴ Expression = (tan 1°.tan 89°) (tan 2°.tan 88°) ---- tan 45°
= (tan 1°.cot 1°). (tan 2°.cot 2°) ---- tan 45°
= 1.1 ----1 = 1Correct Option: A
tan (90° – θ) = cotθ tanθ.cotθ = 1
tan 89° = tan (90° – 1°) = cot 1°.
tan 88° = tan (90° – 2°) = cot 2°.
∴ Expression = (tan 1°.tan 89°) (tan 2°.tan 88°) ---- tan 45°
= (tan 1°.cot 1°). (tan 2°.cot 2°) ---- tan 45°
= 1.1 ----1 = 1
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The numerical value of cos² 45° + cos² 60° - tan² 30° - sin² 30° is sin² 60° sin² 45° cot² 45° cot² 30°
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cos² 45° + cos²60° - tan²30° - sin²30° sin²60° sin²45° cot²45° cot²30° 
= 1 × 4 + 1 × 2 - 1 × 1 - 1 2 3 4 3 4 × 3 = 2 + 1 - 1 - 1 3 2 3 12 = 8 + 6 - 4 - 1 = 9 = 3 12 12 4
Correct Option: B
cos² 45° + cos²60° - tan²30° - sin²30° sin²60° sin²45° cot²45° cot²30° = 1 × 4 + 1 × 2 - 1 × 1 - 1 2 3 4 3 4 × 3 = 2 + 1 - 1 - 1 3 2 3 12 = 8 + 6 - 4 - 1 = 9 = 3 12 12 4
- If sin θ + sin²θ = 1 then cos²θ + cos4θ is equal to
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sinθ + sin²θ = 1
⇒ sinθ = 1 – sin²θ = cos2θ
∴ cos²θ + cos4θ
= cos²θ + (cos²θ)²
= cos²θ + sin²θ = 1Correct Option: B
sinθ + sin²θ = 1
⇒ sinθ = 1 – sin²θ = cos2θ
∴ cos²θ + cos4θ
= cos²θ + (cos²θ)²
= cos²θ + sin²θ = 1
- If x cosθ – sinθ = 1, then x² + (1 +x² ) sinq equals
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x cosθ – sinθ = 1
⇒ x cosθ = 1 + sinθ⇒ x = 1 + sin θ cos θ cos θ
⇒ x = secθ + tanθ --- (i)
∵ sec²θ – tan²θ = 1
⇒ (secθ + tanθ) (secθ – tanθ) =1⇒ secθ – tanθ = 1 (ii) x
From equation (i) + (ii),2secθ = x + 1 = x² + 1 x x ⇒ secθ = x² + 1 2x
From equation (i) – (ii),2tanθ = x – 1 = x² - 1 x x ∴ tanθ = x² - 1 2x ∴ sinθ = tanθ secθ = x² - 1 × 2x = x² - 1 2x x² + 1 x² + 1
∴ Expression = x² – (1 + x² ) sinθ= x² - (1 + x²) × x² - 1 = x² - x² + 1 = 1 x² + 1
Note : In the original equation x² + (1 + x² ) sinθ has been given that seems incorrect
Correct Option: B
x cosθ – sinθ = 1
⇒ x cosθ = 1 + sinθ⇒ x = 1 + sin θ cos θ cos θ
⇒ x = secθ + tanθ --- (i)
∵ sec²θ – tan²θ = 1
⇒ (secθ + tanθ) (secθ – tanθ) =1⇒ secθ – tanθ = 1 (ii) x
From equation (i) + (ii),2secθ = x + 1 = x² + 1 x x ⇒ secθ = x² + 1 2x
From equation (i) – (ii),2tanθ = x – 1 = x² - 1 x x ∴ tanθ = x² - 1 2x ∴ sinθ = tanθ secθ = x² - 1 × 2x = x² - 1 2x x² + 1 x² + 1
∴ Expression = x² – (1 + x² ) sinθ= x² - (1 + x²) × x² - 1 = x² - x² + 1 = 1 x² + 1
Note : In the original equation x² + (1 + x² ) sinθ has been given that seems incorrect
