126. If x sin 45° = y cosec 30°, then	x4	is equal to
     	y4

1. 1. 4³

   
   
   

   2. 6³

   
   
   

   3. 2³

   
   
   

   4. 8³
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   x sin 45° = y cosec 30°
   

   ⇒ x ×	1	= y × 2
   	√2

   
   

   ⇒	x	= 2√2
   	y

   
   

   ⇒	x4	= (2√2)4= 24 × 22
   	y4

   
   = 2⁶ = 4³
   

   Correct Option: A

   x sin 45° = y cosec 30°
   

   ⇒ x ×	1	= y × 2
   	√2

   
   

   ⇒	x	= 2√2
   	y

   
   

   ⇒	x4	= (2√2)4= 24 × 22
   	y4

   
   = 2⁶ = 4³
   

---

127. If cosec θ – cotθ =	7	, the value of cosecθ is :
     	2

1. 1. 	47
      	28

   
   
   

   2. 	51
      	28

   
   
   

   3. 	53
      	28

   
   
   

   4. 	49
      	28

   
   
   

---

1. View Hint View Answer Discuss in Forum

   cosecθ – cotθ =	7	.......(i)
   	2

   
   cosec²θ – cot²θ = 1
   ⇒ (cosecθ + cotθ) (cosecθ – cotθ) = 1
   ⇒ cosecθ + cotθ
   

   =		1	=	2	........(iii)
   		cosecθ - cotθ		7

   
   On adding both equations,
   

   2 cosecθ =		7	+	2
   		2		7

   
   

   =		49 + 4	+	53
   		14		14

   
   

   ⇒ cosecθ =	53
   	28

   
   

   Correct Option: C

   cosecθ – cotθ =	7	.......(i)
   	2

   
   cosec²θ – cot²θ = 1
   ⇒ (cosecθ + cotθ) (cosecθ – cotθ) = 1
   ⇒ cosecθ + cotθ
   

   =		1	=	2	........(iii)
   		cosecθ - cotθ		7

   
   On adding both equations,
   

   2 cosecθ =		7	+	2
   		2		7

   
   

   =		49 + 4	+	53
   		14		14

   
   

   ⇒ cosecθ =	53
   	28

   
   

---

---

128. If tan θ =	3	and θ is acute, then cosec θ
     	4

1. 1. 	4
      	5

   
   
   

   2. 	5
      	3

   
   
   

   3. 2

   
   
   

   4. 	1
      	2

   
   
   

---

1. View Hint View Answer Discuss in Forum

   tan θ =	3
   	4

   
   

   ∴ cot θ =	4
   	3

   
   ∵ cosec² θ – cot² θ = 1
   ⇒ cosec θ = √1 + cot² θ
   
   

   =	5
   	3

   
   

   Correct Option: B

   tan θ =	3
   	4

   
   

   ∴ cot θ =	4
   	3

   
   ∵ cosec² θ – cot² θ = 1
   ⇒ cosec θ = √1 + cot² θ
   
   

   =	5
   	3

   
   

---

129. If tan α = n tan β and sin α = m sin β, then cos² α is

1. 1. 	m²
      	n² + 1

   
   
   

   2. 	m²
      	n²

      
      

   
   
   

   3. 	m² - 1
      	n² - 1

   
   
   

   4. 	m² + 1
      	n² + 1

   
   
   

---

1. View Hint View Answer Discuss in Forum

   tan α = n tan β
   

   ⇒ tan β	1	tan α
   	n

   
   

   ⇒ cot β	n	and
   	tan α

   
   

   sin α = m sin β =	1	sin α
   	m

   
   

   ⇒ cosec β =	m
   	sin α

   
   [∵ cosec²β - cot² β = 1]
   

   ⇒		m²	-	n²	= 1
   		sin²α		sin²α

   
   

   ⇒		m²	-	n²cos²α	= 1
   		sin²α		tan²α

   
   

   ⇒	m² - n²cos²α	= 1
   	sin²α

   
   ⇒ m² - n² cos²α = sin²α
   = 1 - cos²α
   ⇒ m² - 1n² cos²α - cos²α
   

   ⇒cos²α =	m² - 1
   	n² - 1

   
   

   Correct Option: C

   tan α = n tan β
   

   ⇒ tan β	1	tan α
   	n

   
   

   ⇒ cot β	n	and
   	tan α

   
   

   sin α = m sin β =	1	sin α
   	m

   
   

   ⇒ cosec β =	m
   	sin α

   
   [∵ cosec²β - cot² β = 1]
   

   ⇒		m²	-	n²	= 1
   		sin²α		sin²α

   
   

   ⇒		m²	-	n²cos²α	= 1
   		sin²α		tan²α

   
   

   ⇒	m² - n²cos²α	= 1
   	sin²α

   
   ⇒ m² - n² cos²α = sin²α
   = 1 - cos²α
   ⇒ m² - 1n² cos²α - cos²α
   

   ⇒cos²α =	m² - 1
   	n² - 1

   
   

---

---

130. sin²θ– 3 sin θ + 2 = 0 will be true if

1. 1. 0 ≤ θ < 90

   
   
   

   2. 0 < θ < 90

   
   
   

   3. θ = 0°

   
   
   

   4. θ = 90°

   
   
   

---

1. View Hint View Answer Discuss in Forum

   sin² θ – 3 sin θ + 2 = 0
   ⇒ sin² θ – 2 sin θ – sin θ + 2 = 0
   ⇒ sin θ (sin θ – 2) –1 (sin θ – 2) = 0
   ⇒ (sin θ – 1) (sin θ – 2) = 0
   ⇒ sin θ = 1 = sin 90°
   ⇒ θ = 90° and sin θ ≠ 2

   Correct Option: D

   sin² θ – 3 sin θ + 2 = 0
   ⇒ sin² θ – 2 sin θ – sin θ + 2 = 0
   ⇒ sin θ (sin θ – 2) –1 (sin θ – 2) = 0
   ⇒ (sin θ – 1) (sin θ – 2) = 0
   ⇒ sin θ = 1 = sin 90°
   ⇒ θ = 90° and sin θ ≠ 2

---