376. The value of		4	=	3	+ 3 sin² α will be
     		1 + cot²α		1 + cot²α

1. 1. 4

   
   
   

   2. - 1

   
   
   

   3. 2

   
   
   

   4. 3

   
   
   

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   		4	+	1	+ 3 sin²α
   		1 + tan²α		1 + cot²α

   
   

   		4	+	1	+ 3 sin²α
   		sec²α		cosec²α

   
   = 4 cos²α + sin²α + 3 sin²α
   = 4(cos²α + sin²α) = 4

   Correct Option: A

   		4	+	1	+ 3 sin²α
   		1 + tan²α		1 + cot²α

   
   

   		4	+	1	+ 3 sin²α
   		sec²α		cosec²α

   
   = 4 cos²α + sin²α + 3 sin²α
   = 4(cos²α + sin²α) = 4

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377. The numerical value of		1	=	3	+ 2 sin² θ will be
     		1 + cot²θ		1 + tan²θ

1. 1. 2

   
   
   

   2. 5

   
   
   

   3. 6

   
   
   

   4. 3

   
   
   

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1. View Hint View Answer Discuss in Forum

   		1	+	3	+ 2sin² θ
   		1 + cot² θ		1 + tan² θ

   
   =

   		1	+	3	+ 2sin² θ
   		cosec² θ		sec² θ

   
   = sin² θ + 3cos² θ + 2sin² θ
   = 3(sin² θ + cos² θ) = 3
   

   Correct Option: D

   		1	+	3	+ 2sin² θ
   		1 + cot² θ		1 + tan² θ

   
   =

   		1	+	3	+ 2sin² θ
   		cosec² θ		sec² θ

   
   = sin² θ + 3cos² θ + 2sin² θ
   = 3(sin² θ + cos² θ) = 3
   

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378. If x cos θ – y sin θ = √x² + y²		cos²θ	+	sin²θ	=	1	, then the correct relation is
     		a²		b²		x² + y²

1. 1. 		x²	-	y²	= 1
      		b²		a²

   
   
   

   2. 		x²	+	y²	= 1
      		a²		b²

   
   
   

   3. 		x²	+	y²	= 1
      		b²		a²

   
   
   

   4. 		x²	-	y²	= 1
      		a²		b²

   
   
   

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1. View Hint View Answer Discuss in Forum

   According to question,
   x cosθ – y sinθ = √x² + y²...(i)
   

   	cos²θ	+	sin²θ	=	1	...(ii)
   	a²		b²		x² + y²

   
   
   

   sinθ =	- y
   	√x² + y²

   
   

   cosθ =	x
   	√x² + y²

   
   From equation (i)
   

   	x	cosθ -	y	sinθ = 1
   	√x² + y²		√x² + y²

   
   

   ∴	cos²θ	+	sin²θ	=	1
   	a²		b²		x² + y²

   
   

   ⇒	x²	+	y²	=	1
   	(x² + y²)a²		(x² + y²)b²		x² + y²

   
   

   ⇒	x²	+	y²	= 1
   	a²		b²

   
   

   Correct Option: B

   According to question,
   x cosθ – y sinθ = √x² + y²...(i)
   

   	cos²θ	+	sin²θ	=	1	...(ii)
   	a²		b²		x² + y²

   
   
   

   sinθ =	- y
   	√x² + y²

   
   

   cosθ =	x
   	√x² + y²

   
   From equation (i)
   

   	x	cosθ -	y	sinθ = 1
   	√x² + y²		√x² + y²

   
   

   ∴	cos²θ	+	sin²θ	=	1
   	a²		b²		x² + y²

   
   

   ⇒	x²	+	y²	=	1
   	(x² + y²)a²		(x² + y²)b²		x² + y²

   
   

   ⇒	x²	+	y²	= 1
   	a²		b²

   
   

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379. If cos x + cos²x = 1, the numerical value of (sin¹²x + 3 sin¹⁰x + 3 sin⁸x + sin⁶x – l) is :

1. 1. - 1

   
   
   

   2. 2

   
   
   

   3. 0

   
   
   

   4. 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   cos x + cos² x = 1
   ⇒ cos x = 1 – cos² x = sin² x ...(i)
   ∴ sin¹²x + 3 sin¹⁰x + 3 sin⁸x + sin⁶x – 1
   = (sin⁴ x + sin² x)³ – 1
   = (cos²x + sin²x)³ – 1 [By (i)]
   = 1 – 1 = 0

   Correct Option: C

   cos x + cos² x = 1
   ⇒ cos x = 1 – cos² x = sin² x ...(i)
   ∴ sin¹²x + 3 sin¹⁰x + 3 sin⁸x + sin⁶x – 1
   = (sin⁴ x + sin² x)³ – 1
   = (cos²x + sin²x)³ – 1 [By (i)]
   = 1 – 1 = 0

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380. If		sec θ + tan θ	=	5	, then sin θ is equal to
     		sec θ - tan θ		3

1. 1. 1

      4

   
   
   

   2. 1

      3

   
   
   

   3. 2

      3

   
   
   

   4. 3

      4

   
   
   

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1. View Hint View Answer Discuss in Forum

   		secθ + tanθ	=	5
   		secθ - tanθ		3

   
   ⇒ 5 sec θ – 5 tan θ
   = 3 sec θ + 3 tan θ
   ⇒ 2 sec θ = 8 tan θ
   

   =		tanθ	=	2	=	1
   		secθ		8		4

   
   

   ⇒		sinθ	× cosθ =	1
   		cosθ		4

   
   

   ⇒ sinθ =	1
   	4

   
   

   Correct Option: A

   		secθ + tanθ	=	5
   		secθ - tanθ		3

   
   ⇒ 5 sec θ – 5 tan θ
   = 3 sec θ + 3 tan θ
   ⇒ 2 sec θ = 8 tan θ
   

   =		tanθ	=	2	=	1
   		secθ		8		4

   
   

   ⇒		sinθ	× cosθ =	1
   		cosθ		4

   
   

   ⇒ sinθ =	1
   	4

   
   

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