Trigonometry
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Evaluate : tan²60° + 4cos²45° + 3sec²30° + 5cos²90° cosec 30° + sec 60° - cotsec²30°
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tan² 60° + 4cos² 45° + 3sec² 30° + 5cos² 90° cosec 30° + sec 60° – cot² 30° = (√3)² + 4 × 
1 
² + 3 2 ² + 5 × 0 √2 √3 2 + 2 - ((√3)² = 3 + 4 × 1 + 3 × 4 + 0 2 3 4 - 3
= 3 + 2 + 4 = 9Correct Option: B
tan² 60° + 4cos² 45° + 3sec² 30° + 5cos² 90° cosec 30° + sec 60° – cot² 30° = (√3)² + 4 × 1 ² + 3 2 ² + 5 × 0 √2 √3 2 + 2 - ((√3)² = 3 + 4 × 1 + 3 × 4 + 0 2 3 4 - 3
= 3 + 2 + 4 = 9
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If sin(x + y) = a + b , then the value of tan x is equal to sin(x - y) a - b tan y
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Here,
sin(x + y) = a + b sin(x - y) a - b
Using componendo and dividendo both sides we get,sin(x + y) + sin(x - y) sin(x + y) - sin(x - y) = a + b + a - b a + b - (a - b) = 2sin (x + y + x - y) cos (x + y - x + y) 2 2 2sin (x + y - x + y) cos (x + y + x - y) 2 2 = 2a 2a
∵ sinC + sinD= 2sin C + D cos C - D sinC - sinD 2 2 = 2cos C + D sin C - D 2 2 ⇒ sinx.cosy = a siny.cosx b ⇒ tan x = a tan y b Correct Option: B
Here,
sin(x + y) = a + b sin(x - y) a - b
Using componendo and dividendo both sides we get,sin(x + y) + sin(x - y) sin(x + y) - sin(x - y) = a + b + a - b a + b - (a - b) = 2sin (x + y + x - y) cos (x + y - x + y) 2 2 2sin (x + y - x + y) cos (x + y + x - y) 2 2 = 2a 2a
∵ sinC + sinD= 2sin C + D cos C - D sinC - sinD 2 2 = 2cos C + D sin C - D 2 2 ⇒ sinx.cosy = a siny.cosx b ⇒ tan x = a tan y b
- 2 (sin6θ + cos6θ) – 3 (sin4θ + cos4θ) + 1 = ?
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2 (sin6θ + cos6θ) – 3 (sin4θ + cos4θ) + 1
= 2[(sin²θ)3 + (cos²θ)3] – 3 [(sin²θ)2 + (cos²θ)2] + 1
= 2 (sin²θ + cos²θ) (sin4θ + cos4θ – sin²θ . cos²θ) – 3 [(sin²θ + cos²θ)² – 2sin²θ . cos²θ] + 1
[∵ a³ + b³ = (a + b) (a² – ab + b²) ; a² + b² = (a + b)² – 2ab]
= 2 sin4θ + 2 cos4θ – 2sin²θ . cos²θ – 3 + 6 sin²θ . cos²θ + 1
= 2 [(sin²θ + cos²θ)2 – 2 sin²θ . cos²θ] – 3 + 4 sin²θ . cos²θ + 1
= 2 – 4 sin²θ . cos²θ – 3 + 4 sin²θ cos²θ + 1 = 2 – 3 + 1
= 0Correct Option: B
2 (sin6θ + cos6θ) – 3 (sin4θ + cos4θ) + 1
= 2[(sin²θ)3 + (cos²θ)3] – 3 [(sin²θ)2 + (cos²θ)2] + 1
= 2 (sin²θ + cos²θ) (sin4θ + cos4θ – sin²θ . cos²θ) – 3 [(sin²θ + cos²θ)² – 2sin²θ . cos²θ] + 1
[∵ a³ + b³ = (a + b) (a² – ab + b²) ; a² + b² = (a + b)² – 2ab]
= 2 sin4θ + 2 cos4θ – 2sin²θ . cos²θ – 3 + 6 sin²θ . cos²θ + 1
= 2 [(sin²θ + cos²θ)2 – 2 sin²θ . cos²θ] – 3 + 4 sin²θ . cos²θ + 1
= 2 – 4 sin²θ . cos²θ – 3 + 4 sin²θ cos²θ + 1 = 2 – 3 + 1
= 0
- If sinθ + sin²θ + sin³θ = 1, then, cos6θ – 4 cos4θ + 8 cos²θ = ?
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sinθ + sin²θ + sin³θ = 1
⇒ sinθ + sin³θ = 1 – sin²θ
⇒ sinθ (1 + sin²θ) = cos²θ
⇒ sin²θ (1 + sin²θ)² = cos4θ
⇒ (1 – cos²θ) {1 + (1 – cos²θ)}2= cos4θ
⇒ (1 – cos²θ) (2 – cos²θ)² = cos4θ
⇒ (1 – cos²θ) (4 – 4cos²θ + cos4θ) = cos4θ
⇒ 4 – 4cos²θ + cos4θ – 4cos²θ + 4cos4θ – cos6θ = cos4θ
⇒ –cos6θ + 4 cos4θ – 8 cos²θ + 4 = 0
⇒ cos6θ – 4cos4θ + 8cos²θ = 4Correct Option: C
sinθ + sin²θ + sin³θ = 1
⇒ sinθ + sin³θ = 1 – sin²θ
⇒ sinθ (1 + sin²θ) = cos²θ
⇒ sin²θ (1 + sin²θ)² = cos4θ
⇒ (1 – cos²θ) {1 + (1 – cos²θ)}2= cos4θ
⇒ (1 – cos²θ) (2 – cos²θ)² = cos4θ
⇒ (1 – cos²θ) (4 – 4cos²θ + cos4θ) = cos4θ
⇒ 4 – 4cos²θ + cos4θ – 4cos²θ + 4cos4θ – cos6θ = cos4θ
⇒ –cos6θ + 4 cos4θ – 8 cos²θ + 4 = 0
⇒ cos6θ – 4cos4θ + 8cos²θ = 4
- The shadow of a vertical tower increases 10 metre, when the altitude of the sun changes from 45° to 30°. What is the height of tower ? (π = 1.73)
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AB = Tower = h Metre
CD = 10 metre, AC = x metre
(let)
∠BCA = 45°, ∠BDA = 30°
In ∆ACB,tan 45° = AB AC ⇒ 1 = h x
⇒ h = x --- (i)
In ∆DAB,tan 30° = AB AD ⇒ 1 = h √x + 10 7
⇒ x + 10 = √3h
⇒ h + 10 = √3h
⇒ h (√3 – 1) = 10⇒ h = 10 √3 – 1 = 10 × √3 + 1 √3 – 1 √3 + 1 = 10(√3 + 1) = 5(1.73 + 1) 2
= 13.65 metreCorrect Option: B
AB = Tower = h Metre
CD = 10 metre, AC = x metre
(let)
∠BCA = 45°, ∠BDA = 30°
In ∆ACB,tan 45° = AB AC ⇒ 1 = h x
⇒ h = x --- (i)
In ∆DAB,tan 30° = AB AD ⇒ 1 = h √x + 10 7
⇒ x + 10 = √3h
⇒ h + 10 = √3h
⇒ h (√3 – 1) = 10⇒ h = 10 √3 – 1 = 10 × √3 + 1 √3 – 1 √3 + 1 = 10(√3 + 1) = 5(1.73 + 1) 2
= 13.65 metre
