501. If tan θ = tan 30° . tan 60° and q is an acute angle, then 2θ is equal to

1. 1. 30°

   
   
   

   2. 45°

   
   
   

   3. 90°

   
   
   

   4. 0°

   
   
   

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1. View Hint View Answer Discuss in Forum

   tanθ = tan30° . tan 60°
   

   ⇒  tanθ =	1	× √3 = 1
   	√3

   
   ⇒  tanθ = tan45°
   ⇒  θ = 45°
   ∴  2θ = 2 × 45° = 90°

   Correct Option: C

   tanθ = tan30° . tan 60°
   

   ⇒  tanθ =	1	× √3 = 1
   	√3

   
   ⇒  tanθ = tan45°
   ⇒  θ = 45°
   ∴  2θ = 2 × 45° = 90°

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502. If sec 15θ = cosec 15θ (0° < θ < 10°) then value of θ is

1. 1. 9°

   
   
   

   2. 5°

   
   
   

   3. 8°

   
   
   

   4. 3°

   
   
   

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1. View Hint View Answer Discuss in Forum

   sec15θ = cosec15θ
   ⇒  sec15θ = sec (90° – 15θ)
   ⇒  15θ = 90° – 15θ
   ⇒  15θ = 90° – 15θ
   ⇒  15θ + 15θ = 90°
   ⇒  30θ = 90°
   

   ⇒  θ =	90°
   	30

   
   = 3°

   Correct Option: D

   sec15θ = cosec15θ
   ⇒  sec15θ = sec (90° – 15θ)
   ⇒  15θ = 90° – 15θ
   ⇒  15θ = 90° – 15θ
   ⇒  15θ + 15θ = 90°
   ⇒  30θ = 90°
   

   ⇒  θ =	90°
   	30

   
   = 3°

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503. If x = a cos θ cos θ, y = a cosθ sinθ and z = a sinθ, then the value of (x² + y² + z²) is

1. 1. 2a²

   
   
   

   2. 4a²

   
   
   

   3. 9a²

   
   
   

   4. a²

   
   
   

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1. View Hint View Answer Discuss in Forum

   x = a cosθ. cosφ
   y = a cosθ. sinφ
   z = a sinθ
   ∴  x² + y² + z²
   = a²cos². cos²θ + a²cos².sin²φ + a²sin²θ
   = a²cos²θ(cos²φ + sin²φ) + a²sin²θ
   = a²cos²θ + a²sin²θ
   = a²(cos²θ + sin²θ) = a²

   Correct Option: D

   x = a cosθ. cosφ
   y = a cosθ. sinφ
   z = a sinθ
   ∴  x² + y² + z²
   = a²cos². cos²θ + a²cos².sin²φ + a²sin²θ
   = a²cos²θ(cos²φ + sin²φ) + a²sin²θ
   = a²cos²θ + a²sin²θ
   = a²(cos²θ + sin²θ) = a²

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504. If   5 sin2θ + 4cos2θ =	9	and 0 < θ <	π	then tanθ is equal to
     	2		2

1. 1. 1

   
   
   

   2. 0

   
   
   

   3. –1

   
   
   

   4. 1

      4

   
   
   

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1. View Hint View Answer Discuss in Forum

   5 sin2θ + 4cos2θ =	9
   	2

   
   ⇒  10 sin²θ + 8 cos²θ = 9
   On dividing by cos²θ
   

   10 sin2θ	+	8 cos2θ	=	9	= 9 sec2θ
   cos2θ		cos2θ		cos2θ

   
   ⇒  10 tan²θ + 8 = 9 (1 + tan²θ)
   ⇒  10 tan²θ + 8 = 9 + 9 tan²θ
   ⇒  10 tan²θ – 9 tan²θ = 9 – 8
   ⇒  tan²θ = 1 ⇒ tan θ = ± 1
   

   ∵  0 < θ <	π	,   ∴  tanθ = 1
   	2

   Correct Option: A

   5 sin2θ + 4cos2θ =	9
   	2

   
   ⇒  10 sin²θ + 8 cos²θ = 9
   On dividing by cos²θ
   

   10 sin2θ	+	8 cos2θ	=	9	= 9 sec2θ
   cos2θ		cos2θ		cos2θ

   
   ⇒  10 tan²θ + 8 = 9 (1 + tan²θ)
   ⇒  10 tan²θ + 8 = 9 + 9 tan²θ
   ⇒  10 tan²θ – 9 tan²θ = 9 – 8
   ⇒  tan²θ = 1 ⇒ tan θ = ± 1
   

   ∵  0 < θ <	π	,   ∴  tanθ = 1
   	2

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505. Which one of the following is true for 0°<θ<90°

1. 1. cosθ > cos²θ

   
   
   

   2. cosθ < cos²θ

   
   
   

   3. cosθ ≥ cos²θ

   
   
   

   4. cosθ ≤ cos²θ

   
   
   

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1. View Hint View Answer Discuss in Forum

   For 0° < q < 90°
   cosθ > cos²θ because cosθ° = 1
   and cos 90° = 0

   Correct Option: A

   For 0° < q < 90°
   cosθ > cos²θ because cosθ° = 1
   and cos 90° = 0

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