221. If 5 cos θ + 12 sin θ = 13, 0° < θ < 90°, then the value of sin q is

1. 1. 	5
      	13

   
   
   

   2. -	12
      	13

   
   
   

   3. 	6
      	13

   
   
   

   4. 	12
      	13

   
   
   

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1. View Hint View Answer Discuss in Forum

   5 cosθ + 12 sinθ = 13
   

   ⇒		5	cosθ +	12	sin θ = 1
   		13		13

   
   ∵ sin²θ + cos²θ = 1
   

   ∴ sinθ =		12	. cosθ =	5
   		13		13

   
   

   Correct Option: D

   5 cosθ + 12 sinθ = 13
   

   ⇒		5	cosθ +	12	sin θ = 1
   		13		13

   
   ∵ sin²θ + cos²θ = 1
   

   ∴ sinθ =		12	. cosθ =	5
   		13		13

   
   

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222. If 7sin²θ + 3cos²θ = 4, then the value of tan θ is (θ is acute)

1. 1. 	1
      	√3

   
   
   

   2. 	1
      	√2

   
   
   

   3. √3

   
   
   

   4. 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   7		sin²θ	+	3cos²θ	=	4
   		cos²θ		cos²θ		cos²θ

   
   ⇒ 7 tan²θ + 3 = 4 sec²θ
   ⇒ 7 tan²θ + 3 = 4 (1 + tan²θ)
   ⇒ 7 tan²θ + 3 = 4 + 4 tan²θ
   ⇒ 7 tan²θ – 4 tan²θ = 4 – 3
   

   ⇒ 4 tan²θ = 1 ⇒ tan²θ =	1
   	3

   
   

   ⇒ tanθ =	1
   	√3

   Correct Option: A

   7		sin²θ	+	3cos²θ	=	4
   		cos²θ		cos²θ		cos²θ

   
   ⇒ 7 tan²θ + 3 = 4 sec²θ
   ⇒ 7 tan²θ + 3 = 4 (1 + tan²θ)
   ⇒ 7 tan²θ + 3 = 4 + 4 tan²θ
   ⇒ 7 tan²θ – 4 tan²θ = 4 – 3
   

   ⇒ 4 tan²θ = 1 ⇒ tan²θ =	1
   	3

   
   

   ⇒ tanθ =	1
   	√3

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223. The value of (cosec a – sin a) (sec a – cos a) (tan a + cot a) is

1. 1. 1

   
   
   

   2. 6

   
   
   

   3. 2

   
   
   

   4. 4

   
   
   

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1. View Hint View Answer Discuss in Forum

   Expression
   = (cosec a – sin a) (sec a – cos a) (tan a + cot a)
   
   

   	sin²a + cos²a
   	cosa . sina

   
   

   =		cos²a	×	sin²a	×	1
   		sina		cosa		cosa.sina

   
   = 1

   Correct Option: A

   Expression
   = (cosec a – sin a) (sec a – cos a) (tan a + cot a)
   
   

   	sin²a + cos²a
   	cosa . sina

   
   

   =		cos²a	×	sin²a	×	1
   		sina		cosa		cosa.sina

   
   = 1

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224. If sin A + sin²A = 1, then the value of cos²A + cos₄A is
     

1. 1. 2

   
   
   

   2. 1	2
      	3

   
   
   

   3. 1	2
      	2

   
   
   

   4. 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   sin A + sin²A = 1
   ⇒ sin A = 1 – sin²A = cos²A
   ∴ cos²A + cos⁴A
   = cos²A + (cos²A)²
   = cos²A + sin²A = 1

   Correct Option: D

   sin A + sin²A = 1
   ⇒ sin A = 1 – sin²A = cos²A
   ∴ cos²A + cos⁴A
   = cos²A + (cos²A)²
   = cos²A + sin²A = 1

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225. If cos x + cos²x = 1, then sin⁸x + 2 sin⁶x + sin⁴x is equal to

1. 1. 0

   
   
   

   2. 3

   
   
   

   3. 2

   
   
   

   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)
   ∴ sin⁸ x + 2 sin⁶ x + sin⁴ x
   = (sin⁴ x + sin² x)²
   = ((cos x)² + sin² x)²
   = (cos²x + sin² x)²
   = 1

   Correct Option: D

   cos x + cos² x = 1
   ⇒ cos x = 1 – cos² x
   = sin² x ....... (i)
   ∴ sin⁸ x + 2 sin⁶ x + sin⁴ x
   = (sin⁴ x + sin² x)²
   = ((cos x)² + sin² x)²
   = (cos²x + sin² x)²
   = 1

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