Trigonometry
- 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
- If sin θ – cos θ = (7 / 13) and 0 < θ < 90°, then the value of sin θ + cos θ is
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sin θ – cos θ = 7 ......(i) 13
sin θ + cos θ = x .....(ii)
On squaring both equations and adding,2(sin²θ + cos²θ) = 49 + x² 169 ⇒ x² = 2 - 49 = 338 - 49 169 169 = 289 ⇒ x = 17 169 13
Correct Option: A
sin θ – cos θ = 7 ......(i) 13
sin θ + cos θ = x .....(ii)
On squaring both equations and adding,2(sin²θ + cos²θ) = 49 + x² 169 ⇒ x² = 2 - 49 = 338 - 49 169 169 = 289 ⇒ x = 17 169 13
- If θ be acute and tan θ + cot θ = 2, then the value of tan5 θ + cot10 θ is
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tan θ + cot θ = 2
⇒tan θ + 1 = 2 tan θ
⇒ tan² θ + 1 = 2tan θ
⇒ tan² θ – 2tan θ + 1 = 0
⇒ (tan θ – 1)² = 0
⇒ tan θ = 1 ⇒ cot θ = 1
∴ tan5 θ + cot10 θ = 1 + 1 = 2Correct Option: B
tan θ + cot θ = 2
⇒tan θ + 1 = 2 tan θ
⇒ tan² θ + 1 = 2tan θ
⇒ tan² θ – 2tan θ + 1 = 0
⇒ (tan θ – 1)² = 0
⇒ tan θ = 1 ⇒ cot θ = 1
∴ tan5 θ + cot10 θ = 1 + 1 = 2