Bank Math Master

Question: = ? 

Solution:
1/{tanA√(1 - sin²A)} 
= 1/(tanA × √cos²A)
= 1/(tanA × cosA)
= 1/{(sinA/cosA) × cosA}
= 1/sinA
= cosecA

Question: If sinθ + cosθ = √2 sin(90° - θ), then what is the value of tanθ?

Solution:
Given,
 sinθ + cosθ = √2 sin(90° - θ)
⇒ sinθ + cosθ = √2 cosθ
⇒ (sinθ + cosθ)/cosθ = √2 
⇒ (sinθ/cosθ) + (cosθ/cosθ) = √2 
⇒ tanθ + 1 = √2
∴ tanθ = √2 - 1

Question: If tan3A = √3, then what is the value of A?

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

Question: If cos⁴θ - sin⁴θ = 2/3 then the value of (1 - 2sin²θ) is - 

Solution:
cos⁴θ - sin⁴θ = 2/3
⇒ (cos²θ - sin²θ) (cos²θ + sin²θ) = 2/3
⇒ cos²θ - sin²θ = 2/3
⇒ 1 - sin²θ - sin²θ = 2/3
∴ 1 - 2sin²θ = 2/3

Question: The numerical value of is?

Solution:

Question: If tanθ = 1 then sinθ - cos(- θ) = ?

Solution:
Given, 
tanθ = 1
⇒ sinθ/cosθ = 1
⇒ sinθ = cosθ

∴ sinθ - cos(- θ)
= sinθ - cosθ
= cosθ - cosθ
= 0

Question: The angle of elevation at the top of a tower at a point on the ground is 30° at a distance of 75 metre from the foot. Find the height of the tower.



Let the height of the tower is AB = h meter. The angle of elevation at C from the foot of the tower BC = 75 metre of A on the ground is ∠ACB = 30°

From triangle ABC
∴ tan∠ACB = AB/BC
⇒ tan30° = AB/75
⇒ 1/√3 = h/75
⇒ h = 75/√3
⇒ h = 75√3/3
∴ h = 25√3

Question: If θ is a positive acute angle and 4cos²θ - 1 = 0, then the value of tan(θ - 15°) is equal to?

Solution:
Given,
4cos²θ - 1 = 0
⇒ 4cos²θ = 1
⇒ cos²θ = 1/4
⇒ cosθ = 1/2
⇒ cosθ = cos60°
∴ θ = 60°

Now, 
tan(θ - 15°) = tan(60° - 15°)
= tan 45°
= 1

Question: Find the value of

Solution:

Question: The value of is?

Solution:

Question:

Solution:
Given,
sin(A + 18°) = 1/2
⇒ sin(A + 18°) = sin30°
⇒ A + 18° = 30°
⇒ A = 30° - 18°
∴ A = 12° 

Question: sec²17° - (1/tan²73°) - sin17°sec73°

Solution:
sec²17° - (1/tan²73°) - sin17°sec73°
= sec²17° - cot²73° - sin17°sec(90° - 17°)
= sec²17° - cot²(90° - 17°) - sin17°cosec17°
= sec²17° - tan²17° - 1
= 1 - 1 [sec²θ - tan²θ = 1]
= 0

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

Solution:
Given that
secA + tanA = 5/2
Or, 1/(secA + tanA) = 2/5
Or, (sec²A - tan²A)/(secA + tanA) = 2/5
Or, (secA + tanA)(secA - tanA)/(secA + tanA) = 2/5
∴ secA - tanA = 2/5

Question: What is the solution of 2cos²θ + 3sinθ - 3 = 0; where θ is an acute angle.

Solution:
2cos²θ + 3sinθ - 3 = 0
⇒ 2(1 - sin²θ) + 3sinθ - 3 = 0
⇒ 2{(1 + sinθ) (1 - sinθ)} - 3(1 - sinθ) = 0
⇒ (1 - sinθ) {2(1 + sinθ) - 3} = 0
⇒ (1 - sinθ) (2sinθ - 1) = 0

হয়,
1 - sinθ = 0
⇒ sinθ = 1
⇒ sinθ = sin90°
θ = 90°

Or,
2sinθ - 1 = 0
⇒ sinθ = 1/2
⇒ sinθ = ‍sin30°
∴ θ = 30°

θ is an acute angle, ∴ θ = 30°

Question: is equal to?

Solution:
2secθ

Question: What is the length of AB?


Solution:
Given, AB = 8 m and ∠ACB = 60°.

From ΔABC,
Sin60° = AB/AC
Or, √3/2 = AB/8
Or, 2AB = 8√3
∴ AB = 4√3

∴ The length of AB = 4√3 m.

Question: If xtan60° + cos45° = sec45°, then the value of x² + 1 is?

Solution:
xtan60° + cos45° = sec45°
⇒ x . √3 + (1/√2) = √2
⇒ √6x + 1 = 2
⇒ √6x = 1
⇒ x = 1/√6
⇒ x² = (1/√6)²
⇒ x² = 1/6
⇒ x² + 1 = (1/6) + 1
∴ x² + 1 = 7/6

Question: What is the solution of

Solution: