Trigonometry
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Evaluate : sec 39° + 2 (tan 17° . tan 38° . tan 60° . tan 52° . tan 73° – 3 (sin231° + sin259°) cosec 51° √3
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sec 39° + 2 tan17°. tan38° . tan60° . cot (90° – 52°) . cot (90° – 73°) – 3 (sin²31° + cos²(90° – 59°) sec(90° - 51°) √3 = sec 39° + 2 tan17°. tan38° × √3 . cot 38° . cot 17° – 3 (sin²31° + cos² 31°) sec 39° √3 = 1 + 2 × √3(tan 17° . cot17°) .(tan 38° . cot38°) – 3 × 1 = 1 + 2 – 3 = 0 √3
tanq . cotq = 1
tan (90° – θ) cotθ ;
cot (90° – θ) tanθ ;
sec (90° – θ) cosecθCorrect Option: C
sec 39° + 2 tan17°. tan38° . tan60° . cot (90° – 52°) . cot (90° – 73°) – 3 (sin²31° + cos²(90° – 59°) sec(90° - 51°) √3 = sec 39° + 2 tan17°. tan38° × √3 . cot 38° . cot 17° – 3 (sin²31° + cos² 31°) sec 39° √3 = 1 + 2 × √3(tan 17° . cot17°) .(tan 38° . cot38°) – 3 × 1 = 1 + 2 – 3 = 0 √3
tanq . cotq = 1
tan (90° – θ) cotθ ;
cot (90° – θ) tanθ ;
sec (90° – θ) cosecθ
- If cosθ + secθ = 2, then cos5θ + sec5θ= ?
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cosθ + secθ = 2
⇒ cosθ + 1 = 2 cosθ
⇒ cos²θ – 2 cosθ + 1 = 0
⇒ (cosθ – 1)² = 0
⇒ cosθ = 1
∴ cos5θ + sec5θ = 1 + 1 = 2Correct Option: B
cosθ + secθ = 2
⇒ cosθ + 1 = 2 cosθ
⇒ cos²θ – 2 cosθ + 1 = 0
⇒ (cosθ – 1)² = 0
⇒ cosθ = 1
∴ cos5θ + sec5θ = 1 + 1 = 2
- If cosθ – sinθ = 2 sinθ, then cosθ + sinθ = ?
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cosθ – sinθ = 2 sinθ ..(i)
cosθ + sinθ = x ....(ii)
On squaring both the equations and adding.
2 (cos²θ + sin²θ) = 2sin²θ + x²
⇒ x² = 2 – 2 sin²θ = 2 (1 – sin²θ) = 2cos²θ
⇒ x = √2cosθCorrect Option: C
cosθ – sinθ = 2 sinθ ..(i)
cosθ + sinθ = x ....(ii)
On squaring both the equations and adding.
2 (cos²θ + sin²θ) = 2sin²θ + x²
⇒ x² = 2 – 2 sin²θ = 2 (1 – sin²θ) = 2cos²θ
⇒ x = √2cosθ
- The shadow of a vertical tower becomes 30 metres longer when the altitude of the sun changes from 60° to 45°. Find the height of the tower.
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AB = Tower = h metre
DC = 30 metre
BD = x metre
From ∆ABC,tan 45° = AB ⇒ 1 = h ⇒ h = x + 30 ......(i) BC x + 30
From ∆ABD,tan 60° = AB ⇒ √3 = h BD x
h = √3x⇒ x = h .........(ii) √3
∴ h = x + 30⇒ h = h + 30 √3
⇒ (√3 + 1)h = 30√3⇒ h = 30√3 √3 - 1 = 30√3(√3 + 1) (√3 - 1)(√3 + 1)
= 15(3 + √3) metre.Correct Option: B
AB = Tower = h metre
DC = 30 metre
BD = x metre
From ∆ABC,tan 45° = AB ⇒ 1 = h ⇒ h = x + 30 ......(i) BC x + 30
From ∆ABD,tan 60° = AB ⇒ √3 = h BD x
h = √3x⇒ x = h .........(ii) √3
∴ h = x + 30⇒ h = h + 30 √3
⇒ (√3 + 1)h = 30√3⇒ h = 30√3 √3 - 1 = 30√3(√3 + 1) (√3 - 1)(√3 + 1)
= 15(3 + √3) metre.
- If α, β and γ each is positive acute angle, and sin (α + β – γ) = 1/2 , cos (β + γ – α) = 1/2 and tan (γ + α – β) = 1 then 2α + β = ?
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sin(α + β – γ) = 1/2 ,
cos (β + γ– α) = 1/2
and tan(γ + α – β) = 1
⇒ sin (α + β – γ) = sin30°, cos (β + γ– α) = cos60° and tan (γ + α – β) = tan 45°
⇒ α + β – γ = 30° .....(i)
β + γ– α = 60° ......(ii)
γ + α – β = 45° ......(iii)
By equations (i) + (ii) and equations (i) + (iii).
2β = 90° and 2α = 75°
⇒ β = 45° and 2α = 75°
⇒ 2α + β = 75° + 45° = 120°Correct Option: D
sin(α + β – γ) = 1/2 ,
cos (β + γ– α) = 1/2
and tan(γ + α – β) = 1
⇒ sin (α + β – γ) = sin30°, cos (β + γ– α) = cos60° and tan (γ + α – β) = tan 45°
⇒ α + β – γ = 30° .....(i)
β + γ– α = 60° ......(ii)
γ + α – β = 45° ......(iii)
By equations (i) + (ii) and equations (i) + (iii).
2β = 90° and 2α = 75°
⇒ β = 45° and 2α = 75°
⇒ 2α + β = 75° + 45° = 120°
