381. If x = a sec θ cos φ, y = b sec θ sin φ, z = c tan θ, then the value of
     

     		x²	+	y²	-	z²	is :
     		a²		b²		c²

1. 1. 1

   
   
   

   2. 4

   
   
   

   3. 9

   
   
   

   4. 0

   
   
   

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   x = a sec θ. cos φ; y = bsec θ. sin φ, z = c tan θ
   

   ∴		x²	+	y²	-	z²
   		a²		b²		c²

   
   = sec² θ. cos² φ + sec² θ.sin² φ –
   tan² φ
   = sec² θ (cos² φ + sin² φ) – tan² θ
   = sec² θ – tan² θ = 1

   Correct Option: A

   x = a sec θ. cos φ; y = bsec θ. sin φ, z = c tan θ
   

   ∴		x²	+	y²	-	z²
   		a²		b²		c²

   
   = sec² θ. cos² φ + sec² θ.sin² φ –
   tan² φ
   = sec² θ (cos² φ + sin² φ) – tan² θ
   = sec² θ – tan² θ = 1

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382. If		sin θ	=	cos θ	, then sin θ - cos θ is equal to
     		x		y

1. 1. x - y

   
   
   

   2. x + y

   
   
   

   3. x - y

      √x² + y²

   
   
   

   4. y - x

      √x² + y²

   
   
   

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   		sin θ	=	cos θ	=	1
   		x		y		k

   
   ⇒ x ksin θ ; y = kcos θ
   ∴ x² + y²
   = k²(sin²θ + cos²θ)
   ⇒ k = √x² + y²
   ∴ sin θ - cos θ
   

   =		x	-	y	=	x - y
   		k		k		k

   
   

   =	x - y
   	√x² + y²

   
   

   Correct Option: C

   		sin θ	=	cos θ	=	1
   		x		y		k

   
   ⇒ x ksin θ ; y = kcos θ
   ∴ x² + y²
   = k²(sin²θ + cos²θ)
   ⇒ k = √x² + y²
   ∴ sin θ - cos θ
   

   =		x	-	y	=	x - y
   		k		k		k

   
   

   =	x - y
   	√x² + y²

   
   

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383. If tan θ + cot θ = 2, then the value of tanⁿ θ + cotⁿθ (0° < θ < 90°, n is an integer) is

1. 1. 2

   
   
   

   2. 2ⁿ

   
   
   

   3. 2n

   
   
   

   4. 2^{n + 1}

   
   
   

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1. View Hint View Answer Discuss in Forum

   tan θ + cot θ = 2
   

   ⇒ tan θ +	1	= 2
   	tan θ

   
   ⇒ tan⊃ θ - 2tan θ + 1 = 0
   ⇒ (tan θ - 1)² = 0
   ⇒ tanθ = 1 0 ⇒ tanθ = 1
   ∴ cot θ = ⇒ θ= 45°
   ∴ tanⁿ 45° + cotⁿ 45° = 1 + 1 = 2
   

   Correct Option: A

   tan θ + cot θ = 2
   

   ⇒ tan θ +	1	= 2
   	tan θ

   
   ⇒ tan⊃ θ - 2tan θ + 1 = 0
   ⇒ (tan θ - 1)² = 0
   ⇒ tanθ = 1 0 ⇒ tanθ = 1
   ∴ cot θ = ⇒ θ= 45°
   ∴ tanⁿ 45° + cotⁿ 45° = 1 + 1 = 2
   

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384. The value of (1 + cot θ – cosec θ) (1 + tan θ + sec θ) is equal to

1. 1. 1

   
   
   

   2. 2

   
   
   

   3. 0

   
   
   

   4. - 1

   
   
   

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   ( 1 + cot θ - cosec θ) (1+ tan θ + sec θ)
   
   

   =		sin θ + cos θ - 1	×	cos θ + sin θ + 1
   		sin θ		cos θ

   
   

   =	(sin θ + cos θ)² - 1
   	sin θ.cosθ

   
   

   =	sin²θ + cos²θ + 2sinθ.cosθ - 1
   	sin θ.cosθ

   
   

   =	2sinθ.cosθ	= 2
   	sin θ.cosθ

   
   

   Correct Option: B

   ( 1 + cot θ - cosec θ) (1+ tan θ + sec θ)
   
   

   =		sin θ + cos θ - 1	×	cos θ + sin θ + 1
   		sin θ		cos θ

   
   

   =	(sin θ + cos θ)² - 1
   	sin θ.cosθ

   
   

   =	sin²θ + cos²θ + 2sinθ.cosθ - 1
   	sin θ.cosθ

   
   

   =	2sinθ.cosθ	= 2
   	sin θ.cosθ

   
   

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385. If sec θ + tan θ = 2 + √5 , then the value of sin θ + cos θ is :

1. 1. 3

      √5

      
      

   
   
   

   2. √5

   
   
   

   3. 7

      √5

   
   
   

   4. 1

      √5

   
   
   

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   sec θ + tan θ = 2 + √3
   

   ∴ sin θ - tan θ =	1
   	√5 + 2

   
   [∵ sec² θ - tan² θ = 1]
   

   =	√5 - 2	= √5 - 2
   	(√5 + 2)(√5 - 2)

   
   On adding,
   2secθ = 2 + √5 + √5 - 2 = 2√5
   

   ⇒ secθ = √5 ⇒ cosθ =	1
   	√5

   
   On subtracting,
   2tanθ= 2 + √5 - √5 + 2 = 4
   ⇒ tan θ = 2
   

   ∴		tanθ	= sin θ	2
   		sec θ		√5

   
   

   ∴ sin θ + cos θ		2	+	1
   		√5		√5

   
   

   =	3
   	√5

   
   

   Correct Option: A

   sec θ + tan θ = 2 + √3
   

   ∴ sin θ - tan θ =	1
   	√5 + 2

   
   [∵ sec² θ - tan² θ = 1]
   

   =	√5 - 2	= √5 - 2
   	(√5 + 2)(√5 - 2)

   
   On adding,
   2secθ = 2 + √5 + √5 - 2 = 2√5
   

   ⇒ secθ = √5 ⇒ cosθ =	1
   	√5

   
   On subtracting,
   2tanθ= 2 + √5 - √5 + 2 = 4
   ⇒ tan θ = 2
   

   ∴		tanθ	= sin θ	2
   		sec θ		√5

   
   

   ∴ sin θ + cos θ		2	+	1
   		√5		√5

   
   

   =	3
   	√5

   
   

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