366. If sinθ – cosθ = (1 / 2) then value of sinθ + cosθ is :

1. 1. – 2

   
   
   

   2. ± 2

   
   
   

   3. √7

      2

   
   
   

   4. 2

   
   
   

---

1. View Hint View Answer Discuss in Forum

   sinθ – cosθ =	1
   	2

   
   sinθ + cosθ = x
   On squaring and adding,
   

   2(sin²θ + cos²θ) =	1	+ x²
   	4

   
   

   ⇒ x² = 2 -	1	=	7
   	4		4

   
   

   ⇒ x =	√7
   	2

   
   

   Correct Option: C

   sinθ – cosθ =	1
   	2

   
   sinθ + cosθ = x
   On squaring and adding,
   

   2(sin²θ + cos²θ) =	1	+ x²
   	4

   
   

   ⇒ x² = 2 -	1	=	7
   	4		4

   
   

   ⇒ x =	√7
   	2

   
   

---

367. The value of		1	-	1	is
     		(1 + tan² θ)		(1 + cot² θ)

1. 1. 1

      4

   
   
   

   2. 1

   
   
   

   3. 5

      4

   
   
   

   4. 4

      3

      
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Expression
   

   =	1	+	1
   	1 + tan²θ		1 + cot²θ

   
   

   =	1	+	1
   	sec²θ		cosec²θ

   
   = cos²θ + sin²θ = 1
   
   

   Correct Option: B

   Expression
   

   =	1	+	1
   	1 + tan²θ		1 + cot²θ

   
   

   =	1	+	1
   	sec²θ		cosec²θ

   
   = cos²θ + sin²θ = 1
   
   

---

---

368. If cos⁴ θ – sin⁴ θ = (2 / 3) , then the value of 1 – 2 sin²θ is

1. 1. 4

      3

      
      

   
   
   

   2. 0

   
   
   

   3. 2

      3

   
   
   

   4. 1

      3

      
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   cos4θ - sin4θ =	2
   	3

   
   

   ⇒ (cos²θ + sin²θ) (cos²θ - sin²θ =	2
   	3

   
   

   ⇒ 1 - sin²θ - sin²θ =	2
   	3

   
   

   ⇒ 1 - 2sin²θ =	2
   	3

   
   

   Correct Option: C

   cos4θ - sin4θ =	2
   	3

   
   

   ⇒ (cos²θ + sin²θ) (cos²θ - sin²θ =	2
   	3

   
   

   ⇒ 1 - sin²θ - sin²θ =	2
   	3

   
   

   ⇒ 1 - 2sin²θ =	2
   	3

   
   

---

369. If cos θ + sin θ = √2 cos θ, then cos θ – sin θ is

1. 1. √2 tan θ

   
   
   

   2. – √2 cos θ

   
   
   

   3. – √2 sin θ

   
   
   

   4. √2 sin θ
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   cosθ + sinθ = √2cosθ
   On squaring both sides,
   cos²θ + sin²θ + 2cosθ.sinθ= 2cos²θ
   ⇒ cos²θ – sin²θ = 2 sinθ . cosθ
   ⇒ (cosθ + sinθ) (cosθ – sinθ) = 2sinθ. cosθ
   ⇒ √2cosθ (cosθ – sinθ) = 2sinθ . cosθ
   ⇒ cosθ – sinθ
   

   =	2 sinθ . cosθ	= √2sinθ
   	√2cosθ

   
   

   Correct Option: D

   cosθ + sinθ = √2cosθ
   On squaring both sides,
   cos²θ + sin²θ + 2cosθ.sinθ= 2cos²θ
   ⇒ cos²θ – sin²θ = 2 sinθ . cosθ
   ⇒ (cosθ + sinθ) (cosθ – sinθ) = 2sinθ. cosθ
   ⇒ √2cosθ (cosθ – sinθ) = 2sinθ . cosθ
   ⇒ cosθ – sinθ
   

   =	2 sinθ . cosθ	= √2sinθ
   	√2cosθ

   
   

---

---

370. The value of		1	-	1	is
     		cosecθ - cotθ		sinθ

1. 1. 1

   
   
   

   2. cot θ

   
   
   

   3. cosec θ

   
   
   

   4. tan θ

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Expression
   

   =	1	-	1
   	cosecθ - cotθ		sinθ

   
   

   =	cosec²θ - cot²θ	- cosecθ = cosec θ + cotθ – cosecθ = cotθ
   	cosecθ - cotθ

   
   [cosec²θ – cot²θ = 1 & 1/sinθ = cosecθ]
   Method-2 :
   

   1				=	1
   1	-	cosθ			sinθ
   sinθ		sinθ

   
   

   =	sinθ	-	1	-	sin²θ - 1 + cosθ	=	1 - cos²θ - 1 + cosθ
   	1 - cosθ		sinθ		sinθ(1 - cosθ)		sinθ(1 - cosθ)

   
   

   =	cosθ(-cosθ + 1)	=	cosθ	= cotθ
   	sinθ(1 - cosθ)		sinθ

   
   

   Correct Option: B

   Expression
   

   =	1	-	1
   	cosecθ - cotθ		sinθ

   
   

   =	cosec²θ - cot²θ	- cosecθ = cosec θ + cotθ – cosecθ = cotθ
   	cosecθ - cotθ

   
   [cosec²θ – cot²θ = 1 & 1/sinθ = cosecθ]
   Method-2 :
   

   1				=	1
   1	-	cosθ			sinθ
   sinθ		sinθ

   
   

   =	sinθ	-	1	-	sin²θ - 1 + cosθ	=	1 - cos²θ - 1 + cosθ
   	1 - cosθ		sinθ		sinθ(1 - cosθ)		sinθ(1 - cosθ)

   
   

   =	cosθ(-cosθ + 1)	=	cosθ	= cotθ
   	sinθ(1 - cosθ)		sinθ

   
   

---