481. 	2sinθ	simplifies to:
     	cosθ(1 + tan2θ)

     
     

1. 1. cosθ
      

   
   
   

   2. cos2θ
      

   
   
   

   3. sin2θ
      

   
   
   

   4. sinθ

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Expression =	2sinθ
   	cosθ(1 + tan2θ)

   
   

   Expression =	2sinθ
   	cosθ.sec2θ

   
   

   Expression =	2sinθ
   	secθ

   
   [∴ cosθ . secθ = 1]
   Expression = 2 sinθ. cosθ = sin2θ
   

   Correct Option: C

   Expression =	2sinθ
   	cosθ(1 + tan2θ)

   
   

   Expression =	2sinθ
   	cosθ.sec2θ

   
   

   Expression =	2sinθ
   	secθ

   
   [∴ cosθ . secθ = 1]
   Expression = 2 sinθ. cosθ = sin2θ
   

---

482. If sinθ =	5	and θ is acute, what is the value of √(cotθ + tanθ) ?
     	13

1. 1. 2

      √5

      
      

   
   
   

   2. 13

      2√15

      
      

   
   
   

   3. 	-2
      	√5

   
   
   

   4. Cannot be determined

   
   
   

---

1. View Hint View Answer Discuss in Forum

   sinθ =	5
   	13

   
   cosθ = √1 - sin²θ
   cosθ = √{ 1 - (5 / 13)²
   cosθ = √{ 1 - (25 / 169)
   

   cosθ = √	(169 - 25)
   	169

   
   

   cosθ = √	(144)
   	169

   
   

   cosθ =	12
   	13

   
   

   ∴ tanθ =	sinθ	=	5
   	cosθ		12

   
   

   cotθ =	12
   	5

   
   ∴ √(cotθ + tanθ) = √(5 / 12) + (12 / 5)
   

   Required answer = √	(25 + 144)
   	60

   
   

   = √	169
   	60

   
   

   Required answer =	13
   	2 √15

   
   

   Correct Option: B

   sinθ =	5
   	13

   
   cosθ = √1 - sin²θ
   cosθ = √{ 1 - (5 / 13)²
   cosθ = √{ 1 - (25 / 169)
   

   cosθ = √	(169 - 25)
   	169

   
   

   cosθ = √	(144)
   	169

   
   

   cosθ =	12
   	13

   
   

   ∴ tanθ =	sinθ	=	5
   	cosθ		12

   
   

   cotθ =	12
   	5

   
   ∴ √(cotθ + tanθ) = √(5 / 12) + (12 / 5)
   

   Required answer = √	(25 + 144)
   	60

   
   

   = √	169
   	60

   
   

   Required answer =	13
   	2 √15

   
   

---

---

483. If r sinθ = 1, r cosθ = √3 then the value of (r² tanθ) is

1. 1. 4

   
   
   

   2. 1

      √3

   
   
   

   3. 4

      √3

   
   
   

   4. 4√3

   
   
   

---

1. View Hint View Answer Discuss in Forum

   r sinθ = 1        ..... (i)
   r cosθ = √3         ..... (i)
   On squaring and adding both equations,
   r² sin²θ + r²cos²θ = 1 + 3
   ⇒  r² (sin²θ + cos²θ) = 4
   ⇒  r² = 4
   

   Again, =	r sinθ	=	1
   	r cosθ		√3

   
   

   ⇒  tanθ =	1
   	√3

   
   

   ∴  r2 tanθ =	4
   	√3

   Correct Option: C

   r sinθ = 1        ..... (i)
   r cosθ = √3         ..... (i)
   On squaring and adding both equations,
   r² sin²θ + r²cos²θ = 1 + 3
   ⇒  r² (sin²θ + cos²θ) = 4
   ⇒  r² = 4
   

   Again, =	r sinθ	=	1
   	r cosθ		√3

   
   

   ⇒  tanθ =	1
   	√3

   
   

   ∴  r2 tanθ =	4
   	√3

---

484. The value of cos² 20° + cos² 70° is :

1. 1. √2

   
   
   

   2. 2

   
   
   

   3. 1

      √2

   
   
   

   4. 1

   
   
   

---

1. View Hint View Answer Discuss in Forum

   cos220° + cos270°
   = cos2(90° – 70°) + cos270°
   = sin270° + cos270°
   = 1 [cos (90° – θ) = sinq]

   Correct Option: D

   cos220° + cos270°
   = cos2(90° – 70°) + cos270°
   = sin270° + cos270°
   = 1 [cos (90° – θ) = sinq]

---

---

485. If cotθ = 4, then the value of	5sinθ + 3cosθ	is
     	5sinθ − 3cosθ

1. 1. 1

      9

   
   
   

   2. 1

      3

   
   
   

   3. 3

   
   
   

   4. 9

   
   
   

   5. None of these

   
   
   

---

1. View Hint View Answer Discuss in Forum

   cotθ = 4 (Given)
   

   Expression =	5sinθ + 3cosθ
   	5sinθ - 3cosθ

   
   

   =	5	sinθ	+	3cosθ
   		sinθ		sinθ
   	5	sinθ	-	3cosθ
   		sinθ		sinθ

   
   [On dividing numerator and denominator by sinq]
   

   =	5 + 3cotθ
   	5 - 3cotθ

   
   

   =	5 + 3 × 4	=	5 + 12	=	- 17
   	5 - 3 × 4		5 - 12		7

   Correct Option: E

   cotθ = 4 (Given)
   

   Expression =	5sinθ + 3cosθ
   	5sinθ - 3cosθ

   
   

   =	5	sinθ	+	3cosθ
   		sinθ		sinθ
   	5	sinθ	-	3cosθ
   		sinθ		sinθ

   
   [On dividing numerator and denominator by sinq]
   

   =	5 + 3cotθ
   	5 - 3cotθ

   
   

   =	5 + 3 × 4	=	5 + 12	=	- 17
   	5 - 3 × 4		5 - 12		7

---