256. If 7 sin²θ + 3 cos²θ = 4, (0° < θ < 90°), then the value of tan θ is

1. 1. 	1
      	√3

   
   
   

   2. 	1
      	2

   
   
   

   3. 1

   
   
   

   4. √3

   
   
   

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1. View Hint View Answer Discuss in Forum

   7 sin²θ + 3cos²θ = 4
   On dividing both sides by cos²θ 7 tan²θ + 3 = 4 sec²θ
   ⇒ 7 tan²θ + 3 = 4 (1+ tan²θ)
   ⇒ 7 tan²θ + 3 = 4 + 4 tan²θ
   ⇒ 7 tan²θ – 4 tan²θ = 4 – 3
   ⇒ 3 tan²θ = 1
   

   ⇒ tan²θ =	1
   	3

   
   

   ⇒ tanθ =	1
   	√3

   
   

   Correct Option: A

   7 sin²θ + 3cos²θ = 4
   On dividing both sides by cos²θ 7 tan²θ + 3 = 4 sec²θ
   ⇒ 7 tan²θ + 3 = 4 (1+ tan²θ)
   ⇒ 7 tan²θ + 3 = 4 + 4 tan²θ
   ⇒ 7 tan²θ – 4 tan²θ = 4 – 3
   ⇒ 3 tan²θ = 1
   

   ⇒ tan²θ =	1
   	3

   
   

   ⇒ tanθ =	1
   	√3

   
   

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257. If tan 9° = (p / q) , then the value of (sec²81° / 1 + cot²81°) is

1. 1. 	q
      	p

   
   
   

   2. 1

   
   
   

   3. 	p²
      	q²

   
   
   

   4. 	q²
      	p²

   
   
   

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1. View Hint View Answer Discuss in Forum

   ⇒ tan 9° =	p
   	q

   
   

   ∴		sec²81°	=	sec²81°	= 90
   		1 + cot²81°		cosec²81°

   
   

   =	1	× sin²81°
   	cos²81°

   
   = tan²81° = tan² (90° – 9°)
   

   = cot²9° =	q²
   	p²

   
   

   Correct Option: D

   ⇒ tan 9° =	p
   	q

   
   

   ∴		sec²81°	=	sec²81°	= 90
   		1 + cot²81°		cosec²81°

   
   

   =	1	× sin²81°
   	cos²81°

   
   = tan²81° = tan² (90° – 9°)
   

   = cot²9° =	q²
   	p²

   
   

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258. If secθ + tanθ = 5, then the value of tan (tanθ + 1 / tanθ - 1) is

1. 1. 	11
      	7

   
   
   

   2. 	13
      	7

   
   
   

   3. 	15
      	7

   
   
   

   4. 	17
      	7

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   secθ + tanθ = 5
   ∴ sec²θ – tan²θ = 1
   ⇒ (secθ – tanθ) (secθ + tanθ) = 1
   

   ⇒ secθ – tanθ =	1
   	5

   
   ∴ (secθ + tanθ) – (secθ – tanθ)
   

   = 5 -		1	=	25 - 1
   		5		5

   
   

   ⇒ 2tanθ =		24	⇒ tanθ =	12
   		5		5

   
   
   

   =	17
   	7

   
   

   Correct Option: D

   secθ + tanθ = 5
   ∴ sec²θ – tan²θ = 1
   ⇒ (secθ – tanθ) (secθ + tanθ) = 1
   

   ⇒ secθ – tanθ =	1
   	5

   
   ∴ (secθ + tanθ) – (secθ – tanθ)
   

   = 5 -		1	=	25 - 1
   		5		5

   
   

   ⇒ 2tanθ =		24	⇒ tanθ =	12
   		5		5

   
   
   

   =	17
   	7

   
   

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259. If tan²θ = 1 – e² , then the value of secθ + tan³θ cosecθ is

1. 1. (2 + e²)^{3 / 2}

   
   
   

   2. (2 - e²)^{1 / 2}

   
   
   

   3. (2 + e²)^{1 / 2}

   
   
   

   4. (2 - e²)^{3 / 2}

   
   
   

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1. View Hint View Answer Discuss in Forum

   tan²θ = 1 –e²
   ∴ secθ + tan³θ . cosecθ
   = secθ + tan²θ . tanθ . cosecθ
   

   = secθ + tan²θ .		sin θ	.	1
   		cos θ		sin θ

   
   = secθ + tan²θ . secθ
   = secθ .(1 + tan²θ)
   
   

   Correct Option: D

   tan²θ = 1 –e²
   ∴ secθ + tan³θ . cosecθ
   = secθ + tan²θ . tanθ . cosecθ
   

   = secθ + tan²θ .		sin θ	.	1
   		cos θ		sin θ

   
   = secθ + tan²θ . secθ
   = secθ .(1 + tan²θ)
   
   

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260. Which one of the following is true for 0° < q < 90° ?

1. 1. cosθ ≤ cos²θ

   
   
   

   2. cosθ > coscos²θ

   
   
   

   3. cosθ < coscos²θ

   
   
   

   4. cosθ ≥ coscos²θ

   
   
   

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1. View Hint View Answer Discuss in Forum

   When θ = 60°
   

   cosθ =		1	, cos²θ =	1	= 90
   		2		4

   
   ∴ cosθ > cos²θ
   

   Correct Option: B

   When θ = 60°
   

   cosθ =		1	, cos²θ =	1	= 90
   		2		4

   
   ∴ cosθ > cos²θ
   

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