Bank Math Master

Question: If sinθ = cosθ then what is the value of θ?

Solution:
sinθ = cosθ
∴ sinθ/cosθ = 1
⇒ tanθ = 1 
⇒ tanθ = tan45°
∴ θ = 45°

Question: If tanθ = 3/4, then cosecθ = ?

Solution:
এখানে,
tanθ = 3/4 = লম্ব/ভূমি

∴ লম্ব = 3, ভূমি = 4
∴ অতিভুজ = √(3²+ 4²)
= √25 = 5

∴ cosecθ = অতিভুজ/লম্ব
= 5/3

Question: Which trigonometric ratio is undefined in value?

Solution:
sin90° = 1 
cos90° = 0
sec0° = 1
cosec0° = ∞(Undefined)

Question: If A = 45° , then what is the value of (1 - tan²A)/(1 + tan²A)?

Solution:
Here, A = 45°

Now,
(1 - tan²A)/(1 + tan²A)
= {1 - (tan45°)²}/{1 + (tan45°)²}
= (1 - 1²)/(1 + 1²)
= 0/2
= 0

Question: If secA + tanA = 5/2, then what is the value of secA - tanA?

Solution: 
দেয়া আছে,
secA + tanA = 5/2

আমরা জানি,
sec²A - tan²A = 1
⇒ (secA + tanA) (secA - tanA ) =1 
⇒ 5/2(secA - tanA) = 1
∴ (secA - tanA) = 2/5 

Question: If sinθ = 5/13 , then secθ = ?

Solution:
এখানে,
sinθ = 5/13
∴ লম্ব = 5, অতিভুজ = 13

∴ ভূমি = √(13² - 5²) = 12

∴ secθ = অতিভূজ/ভূমি
= 13/12

Question: If θ = 60°, then sec²θ - tan²θ = ?

Solution: 
Given, θ = 60°

Now,
sec²θ - tan²θ
= (sec60°)² - (tan60°)²
= 2² - (√3)²
= 4 - 3
= 1

Question: Find the value of cosec(- π/3) 

Solution:
cosec(- π/3)
= - cosec(π/3)
= - 1/sin(π/3)
= - 1/sin60°
= - 1/(√3/2)
= - 2/√3

Question: find the value of sin²21° + cos²21°

Solution: 
sin²21° + cos²21°
= 1 [sin²θ + cos²θ = 1]

Question: find 

Solution: 

Question: If , what is the value of A?

Solution:

Question: sin(A + 18°) = √3/2, find the value of A.

Solution:
sin(A + 18°) = √3/2
⇒ sin(A + 18°) = sin60°
⇒ A + 18° = 60°
⇒ A = 60° - 18°
∴ A = 42°

Question: rsinθ = 1, rcosθ = √3 then the value of (√3tanθ + 1) = ?

Solution:
rsinθ = 1
rcosθ = √3

Now,
rsinθ/rcosθ = 1/√3
⇒ tanθ = 1/√3
⇒ √3tanθ = 1
⇒ √3tanθ + 1 = 1 + 1
∴ √3tanθ + 1 = 2

Question: If tan(θ - 45°) = 1, then what is the value of sinθ?

Solution:
Given that,
tan(θ - 45°) = 1
⇒ tan(θ - 45°) = tan45°
⇒ (θ - 45°) = 45°
∴  θ = 90°

Now,
sinθ
= sin90°
= 1

Question: If tan3A = √3, then A = ?

Solution:
tan3A = √3
⇒ tan3A = tan60°
⇒ 3A = 60°
⇒ A = 60°/3
∴ A = 20°

Question: What is the value of 1 + {tan²A/(1 + secA)} ?

Solution:
1 + {tan²A/(1 + secA)}
= 1 + {(sce²A - 1)/(1 + secA)}
= {(1 + secA) + (sce²A - 1)}/(1 + secA)
= (1 + secA + sce²A - 1)/(1 + secA)
= (secA + sce²A)/(1 + secA)
= secA(1 + secA)/(1 + secA)
= secA

Question: If sinA + cosA = 1 , then A = ?

Solution:
sinA + cosA = 1
⇒ (sinA + cosA)² = 1²
⇒ sin²A + cos²A + 2sinAcosA = 1
⇒ 1 + 2sinAcosA = 1 
⇒ 2sinAcosA = 1 - 1
⇒ 2sinAcosA = 0
∴ sinAcosA = 0

Here,
sinA = 0
⇒ sinA = sin0°
∴ A = 0°

Or,
cosA = 0
⇒ cosA = cos90°
∴ A = 90°

A = 0°, 90°

Question: Find the greatest value of sin⁴A + cos⁴A.

Solution:
We know,
sin²A + cos²A = 1
⇒(sin²A + cos²A)² = 1²
⇒ (sin²A)² + (cos²A)² + 2sin²Acos²A = 1
⇒ sin⁴A + cos⁴A = 1 - 2sin²Acos²A
⇒ sin⁴A + cos⁴A = 1 - 2sin²90°cos²90° [since we need maximum value]
⇒ sin⁴A + cos⁴A = 1 - 2(1 × 0)
⇒ sin⁴A + cos⁴A = 1 - 0
∴ sin⁴A + cos⁴A = 1