86. If θ is a positive acute angle and tan 2θ tan 3θ = 1, then the value of [2 cos² (5θ / 2) – 1[ is

1. 1. -	1
      	2

      
      

   
   
   

   2. 1

   
   
   

   3. 0

   
   
   

   4. 	1
      	2

   
   
   

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1. View Hint View Answer Discuss in Forum

   tan2θ . tan3θ = 1
   

   ⇒ tan 3θ =	1	= cot 2θ
   	tan 2θ

   
   ⇒ tan3θ = tan (90° – 2θ )
   ⇒ 3θ = 90° – 2θ ⇒ 5θ = 90°
   ⇒ θ = 18
   

   ∴ 2cos ²	5θ	- 1 = 2 cos² 45° - 1
   	2

   
   

   Correct Option: C

   tan2θ . tan3θ = 1
   

   ⇒ tan 3θ =	1	= cot 2θ
   	tan 2θ

   
   ⇒ tan3θ = tan (90° – 2θ )
   ⇒ 3θ = 90° – 2θ ⇒ 5θ = 90°
   ⇒ θ = 18
   

   ∴ 2cos ²	5θ	- 1 = 2 cos² 45° - 1
   	2

   
   

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87. If cos²α + cos²β = 2, then the value of tan&³ α+ sin⁵ β is :

1. 1. - 1

   
   
   

   2. 0

   
   
   

   3. 1

   
   
   

   4. 	1
      	√3

   
   
   

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1. View Hint View Answer Discuss in Forum

   cos²α + cos²β = 2
   ⇒ 1 – sin²α + 1 – sin²β = 2
   ⇒ sin²α + sin²β = 0
   ⇒ sin²α = 0 & sin²β = 0
   ⇒ sinα = sinβ = 0
   ⇒ α = β = 0
   ∴ tan³α + sin⁵β = 0

   Correct Option: B

   cos²α + cos²β = 2
   ⇒ 1 – sin²α + 1 – sin²β = 2
   ⇒ sin²α + sin²β = 0
   ⇒ sin²α = 0 & sin²β = 0
   ⇒ sinα = sinβ = 0
   ⇒ α = β = 0
   ∴ tan³α + sin⁵β = 0

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88. If tan2θ . tan 4θ = 1, then the value of tan 3θis

1. 1. √AAAA
      

   
   
   

   2. 0

   
   
   

   3. 1

   
   
   

   4. 	1
      	√3

   
   
   

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1. View Hint View Answer Discuss in Forum

   tan 2θ =	1	= cot 4θ
   	tan 4θ

   
   ⇒ tan 2θ = tan (90° – 4θ)
   ⇒ 2θ = 90° – 4θ
   ⇒ 6θ = 90° ⇒ θ = 15°
   ∴ tan 3θ = tan 45° = 1

   Correct Option: C

   tan 2θ =	1	= cot 4θ
   	tan 4θ

   
   ⇒ tan 2θ = tan (90° – 4θ)
   ⇒ 2θ = 90° – 4θ
   ⇒ 6θ = 90° ⇒ θ = 15°
   ∴ tan 3θ = tan 45° = 1

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89. If sinθ + cosecθ = 2, then the value of sin⁵θ +cosec⁵θ when 0° ≤ θ ≤ 90°, is

1. 1. 0

   
   
   

   2. 1

   
   
   

   3. 10

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   sinθ + cosecθ = 2
   

   sinθ =	1	= 2
   	sinθ

   
   ⇒ sin² θ - 2sin θ + 1 = 0
   ⇒ (sin θ - 1)² = 0 ⇒ sin θ = 1
   ∴ sin⁵θ + cosec⁵θ = 1 + 1 = 2

   Correct Option: D

   sinθ + cosecθ = 2
   

   sinθ =	1	= 2
   	sinθ

   
   ⇒ sin² θ - 2sin θ + 1 = 0
   ⇒ (sin θ - 1)² = 0 ⇒ sin θ = 1
   ∴ sin⁵θ + cosec⁵θ = 1 + 1 = 2

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90. If A = sin² θ + cos⁴θ, for any value of θ, then the value of A is

1. 1. 1 ≤ A ≤ 2

   
   
   

   2. 	1	≤ A ≤ 1
      	5

   
   
   

   3. 	13	≤ A ≤ 1
      	16

   
   
   

   4. 		3	≤ A ≤	13
      		4		16

   
   
   

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   When θ = 0°
   sin²θ + cos⁴θ = 1
   When θ = 45°,
   
   

   sin²θ + cos4θ =		1	+	1	=	3
   		2		4		4

   
   when θ = 30°,
   
   

   sin²θ + cos4θ =		1	+	9
   		4		16

   
   

   =		4 + 9	=	13
   		16		16

   
   Hence, the value of
   

   A = sin2θ + cos4θ =	13
   	16

   
   

   Correct Option: B

   When θ = 0°
   sin²θ + cos⁴θ = 1
   When θ = 45°,
   
   

   sin²θ + cos4θ =		1	+	1	=	3
   		2		4		4

   
   when θ = 30°,
   
   

   sin²θ + cos4θ =		1	+	9
   		4		16

   
   

   =		4 + 9	=	13
   		16		16

   
   Hence, the value of
   

   A = sin2θ + cos4θ =	13
   	16

   
   

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