Trigonometry
- If 2 (cos²θ– sin²θ) = 1, θ is a positive acute angle, then the value of θ is
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2 (cos⊃θ – sin⊃θ) = 1
⇒ cosec²θ - sin²θ = 1 2 ⇒ 1 - 2 sin²θ = 1 2 ⇒ 2 sin²θ = 1 - 1 2 ⇒ 2 sin²θ = 1 ⇒ sin²θ = 1 2 4 ⇒ sin²θ = ± 1 = sin30° 2 
Correct Option: B
2 (cos⊃θ – sin⊃θ) = 1
⇒ cosec²θ - sin²θ = 1 2 ⇒ 1 - 2 sin²θ = 1 2 ⇒ 2 sin²θ = 1 - 1 2 ⇒ 2 sin²θ = 1 ⇒ sin²θ = 1 2 4 ⇒ sin²θ = ± 1 = sin30° 2
- If cos θ cosec 23° = 1, the value of θ is
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cosθ.cosec23° = 1
⇒ 1 = secθ cos θ
⇒ cosec23° = cosec(90° – θ)
⇒ 23° = 90° – θ
⇒ θ = 90° – 23° = 67°Correct Option: D
cosθ.cosec23° = 1
⇒ 1 = secθ cos θ
⇒ cosec23° = cosec(90° – θ)
⇒ 23° = 90° – θ
⇒ θ = 90° – 23° = 67°
- If sin (3x – 20°) = cos (3y + 20°), then the value of (x + y) is
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sin(3x – 20°)
= cos (3y + 20°)
⇒ sin (3x – 20°)
= sin(90° – 3y – 20°)
= sin(70° – 3y)
⇒ 3x – 20° = 70° – 3y
⇒ 3x + 3y = 90°
⇒ 3(x + y) = 90°
⇒ x + y = 30°Correct Option: B
sin(3x – 20°)
= cos (3y + 20°)
⇒ sin (3x – 20°)
= sin(90° – 3y – 20°)
= sin(70° – 3y)
⇒ 3x – 20° = 70° – 3y
⇒ 3x + 3y = 90°
⇒ 3(x + y) = 90°
⇒ x + y = 30°
- The value of cot θ.tan(90° – θ) – sec(90° – θ) cosecθ + (sin² 25° + sin2 65°) + √3 (tan 5° tan 15°. tan 30°. tan75°. tan 85°) is :
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cotθ. tan (90° – θ) – sec (90° – θ). cosecθ + (sin² 25° + sin² 65°) + √3(tan5°. tan15°. tan30°. tan75°. tan85°)
= cotθ.cotθ – cosecθ. cosecθ + (sin² 25° + cos² 25°) + √3 (tan 5°. cot 5°. tan15°. cot15°. tan30°)
= (cot2θ – cosec2θ) + (sin² 25° + cos² 25°) + √3 × (1 / √3)
= – 1 + 1 + 1 = 1
[sin(90° – θ) = cos θ; cosec² θ – cot² θ = 1; tan (90° – θ) = cotθ ; sec (90° – θ) = cosecθ]Correct Option: A
cotθ. tan (90° – θ) – sec (90° – θ). cosecθ + (sin² 25° + sin² 65°) + √3(tan5°. tan15°. tan30°. tan75°. tan85°)
= cotθ.cotθ – cosecθ. cosecθ + (sin² 25° + cos² 25°) + √3 (tan 5°. cot 5°. tan15°. cot15°. tan30°)
= (cot2θ – cosec2θ) + (sin² 25° + cos² 25°) + √3 × (1 / √3)
= – 1 + 1 + 1 = 1
[sin(90° – θ) = cos θ; cosec² θ – cot² θ = 1; tan (90° – θ) = cotθ ; sec (90° – θ) = cosecθ]
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The value of cot 30° - cot 75° is : tan 15° - tan 60°
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cot 30° = cot (90° – 60°) = tan 60°
cot 75° = cot (90° – 15°) = tan 15°∴ cot 30° - cot 75° tan 15° - tan 60° ∴ tan 60° - tan 15° = - 1 tan 15° - tan 60°
Correct Option: D
cot 30° = cot (90° – 60°) = tan 60°
cot 75° = cot (90° – 15°) = tan 15°∴ cot 30° - cot 75° tan 15° - tan 60° ∴ tan 60° - tan 15° = - 1 tan 15° - tan 60°
