Trigonometry
- If sinθ + cosecθ = 2, then value of sin100θ + cosec100θ is equal to :
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sinθ + cosecθ = 2
⇒ sin θ + 1 = 2 sin θ
⇒ sin2θ – 2 sinθ + 1 = 0
⇒ (sinθ – 1) = 0
⇒ sinθ = 1 ⇒ cosecθ = 1
∴ sin100θ + cosec100θ
= 1+ 1 = 2Correct Option: B
sinθ + cosecθ = 2
⇒ sin θ + 1 = 2 sin θ
⇒ sin2θ – 2 sinθ + 1 = 0
⇒ (sinθ – 1) = 0
⇒ sinθ = 1 ⇒ cosecθ = 1
∴ sin100θ + cosec100θ
= 1+ 1 = 2
- The simplified value of (see x sec y + tan x tan y)² – (sex x tan y + tan x sec y)² is :
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(secx . secy + tanx . tany)² – (sec x . tan y + tan x . sec y)²
= sec² x . sec² y + tan² x . tan²y + 2 sec x . sec y . tan x . tan y – sec² x . tan² y – tan² x . sec²y – 2 sec x . sec y . tan x . tan y
= sec² x . sec²y + tan²x . tan² y – sec² x . tan²y – tan² x . sec²y
= sec²x . sec²y – sec²x . tan²y – tan²x . sec²y + tan² x . tan²y
= sec²x (sec²y – tan²y) – tan²x (sec²y – tan²y)
= sec²x – tan²x = 1Correct Option: D
(secx . secy + tanx . tany)² – (sec x . tan y + tan x . sec y)²
= sec² x . sec² y + tan² x . tan²y + 2 sec x . sec y . tan x . tan y – sec² x . tan² y – tan² x . sec²y – 2 sec x . sec y . tan x . tan y
= sec² x . sec²y + tan²x . tan² y – sec² x . tan²y – tan² x . sec²y
= sec²x . sec²y – sec²x . tan²y – tan²x . sec²y + tan² x . tan²y
= sec²x (sec²y – tan²y) – tan²x (sec²y – tan²y)
= sec²x – tan²x = 1
- If sec²θ + tan²θ = 7, then the value of θ when 0° ≤ θ ≤ 90°, is
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sec²θ + tan²θ = 7
⇒ 1 + tan²θ + tan²θ = 7
⇒ 2 tan²θ = 7 – 1 = 6
⇒ tan²θ = 3 ⇒ tanθ = 3
⇒ θ = 60°Correct Option: A
sec²θ + tan²θ = 7
⇒ 1 + tan²θ + tan²θ = 7
⇒ 2 tan²θ = 7 – 1 = 6
⇒ tan²θ = 3 ⇒ tanθ = 3
⇒ θ = 60°
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If sin θ + cos θ = 3, then the value of sin4θ – cos4θ is sin θ - cos θ
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sin θ + cos θ = 3 sin θ + cos θ
⇒ sinθ + cosθ = 3sinθ – 3 cosθ
⇒ 4cosθ = 2 sinθ ⇒ tanθ = 2
∴ sin4θ – cos4θ
= (sin²θ + cos²θ) (sin²θ – cos²θ)
= sin²θ– cos²θ
= cos²θ (tan² θ – 1)= tan²θ - 1 ²θ = tan²θ - 1 = 4 - 1 = 3 1 + tan²θ 1 + 4 5
Correct Option: C
sin θ + cos θ = 3 sin θ + cos θ
⇒ sinθ + cosθ = 3sinθ – 3 cosθ
⇒ 4cosθ = 2 sinθ ⇒ tanθ = 2
∴ sin4θ – cos4θ
= (sin²θ + cos²θ) (sin²θ – cos²θ)
= sin²θ– cos²θ
= cos²θ (tan² θ – 1)= tan²θ - 1 ²θ = tan²θ - 1 = 4 - 1 = 3 1 + tan²θ 1 + 4 5
- If 2cosθ – sinθ = (1 / √2) , (0° < q < 90°) the value of 2 sinθ + cosθ is
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2 cosθ – sinθ = 1 √2
2sinθ + cosθ = x (Let)
On squaring and adding,
4cos²θ + sin²θ – 4 sinθ . cosθ + 4 sin²θ + cos²θ+ 4 sinθ.cosθ= 1 x² 2 ⇒ 1 + x² = 5 2 ⇒x² = 5 - 1 = 9 ⇒ x = 3 2 2 √2
Correct Option: C
2 cosθ – sinθ = 1 √2
2sinθ + cosθ = x (Let)
On squaring and adding,
4cos²θ + sin²θ – 4 sinθ . cosθ + 4 sin²θ + cos²θ+ 4 sinθ.cosθ= 1 x² 2 ⇒ 1 + x² = 5 2 ⇒x² = 5 - 1 = 9 ⇒ x = 3 2 2 √2
