Bank Math Master

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

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

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

∴ cosθ = ভূমি/অতিভুজ
= 4/5

Question: What is the maximum value of cosθ?

Solution:
cosθ এর সর্বনিম্ন মান - 1 এবং সর্বোচ্চ মান 1
sinθ এর সর্বনিম্ন মান - 1 এবং সর্বোচ্চ মান 1

Question: Which trigonometric ratio is undefined in value?

Solution:
cos90° = 0
sec0° = 1
sin0° = 0
tan90° = ∞(Undefined)

 Question: {(1 - sin²45°)/(1 + sin²45°)} + tan²45° = ?

Solution:
Given that, 
 {(1 - sin²45°)/(1 + sin²45°)} + tan²45°
= {1 - (1/√2)²}/{1 + (1/√2)²} + (1)²   [∴ sin 45° = 1/√2 ও tan 45° = 1] 
= {1 - (1/2)}/{1 + (1/2)} + 1 
= {(2 - 1)/2}/{(2 + 1)/2} + 1 
= (1/2)/(3/2) + 1 
= (1/3) + 1
= (1 + 3)/3
= 4/3

Question: If 1 + tan²θ = 4 and θ < 90°, than what is the value of θ = ?

Solution:
Given that,
1 + tan²θ = 4 and θ < 90°
⇒ sec²θ = 4    ; [sec²θ = 1 + tan²θ]
⇒ (secθ)² = (2)²
⇒ secθ = 2
⇒ secθ = sec60°
⇒ θ = 60°

Question: If B = 45° , then what is the value of (1 - cot²B)/(1 + cot²B)?

Solution:
Here, B = 45°

Now,
(1 - cot²B)/(1 + cot²B)
= {1 - (cot45°)²}/{1 + (cot45°)²}
= (1 - 1²)/(1 + 1²)
= 0/2
= 0

Question: cos²11° + cos²79° = ?

Solution: 
Given that,
cos²11° + cos²79° 
= cos²11° + cos²(90° - 11°)
= cos²11° + sin²11°
= 1

Question: If cosecθ - cotθ = 1/x, then (cosecθ + cotθ) = ?

Solution: 
Given that, 
cosecθ - cotθ = 1/x

We know,
cosec² θ – cot²θ = 1
⇒ (cosecθ + cotθ) (cosecθ - cotθ) = 1
⇒ (cosecθ + cotθ) × (1/x) = 1
⇒ cosecθ + cotθ = x

∴ cosecθ + cotθ = x

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: What is the value of sin60°

Solution:
 sin60° = √3/2
sin30° = 1/2
sin45° = 1/√2
tan30° = 1/√3

Question: (sin⁴θ - cos⁴θ +1) cosec²θ = ?

Solution:
Given that,
(sin⁴θ - cos⁴θ + 1) cosec²θ
= [(sin²θ - cos²θ) (sin²θ + cos²θ) + 1] cosec²θ
= (sin²θ - cos²θ + 1) cosec²θ     ; [sin²A + cos²A = 1]
= [sin²θ - (1 - sin²θ) + 1] cosec²θ
= [2sin²θ - 1 + 1] cosec²θ
= 2sin²θ cosec²θ
= 2sin²θ (1/sin²θ)
= 2

Question: tan³60° - 2sin60° = ?

Solution:
Given that,
tan³60° - 2sin60°
= (√3)³ - 2(√3/2)
= 3√3 - √3
= 2√3

Question: If tan 53° = 4/3, then, what is the value of tan8°?

Solution:
Given that,
tan 53° = 4/3

We know,
tan(A - B) = (tanA - tanB)/(1 + tanA tanB)

Now,
8° = 53° - 45°
tan8° = tan(53° - 45°)
⇒ tan8° = (tan53° - tan45°)/(1 + tan53° tan45°)
⇒ tan8° = {(4/3) - 1}/{1 + (4/3) × 1}
⇒ tan8° = (1/3)/(7/3)
⇒ tan8° = 1/7

Question: If sec²θ + tan²θ = 5/3, then what is the value of tan2θ?

Solution:
We know,
sec²θ = 1 + tan²θ

Given that,
⇒ sec²θ + tan²θ = 5/3
⇒ 1 + tan²θ + tan²θ = 5/3
⇒ 2tan²θ = 2/3
⇒ tanθ = 1/√3
⇒ θ = 30°

∴ tan(2θ) = tan(2 × 30°) = tan60° = √3

Question: (cos²θ + 1/cosec²θ) + 17 = x. What is the value of x²?

Solution:
We know,
sin²θ + cos²θ = 1

Given that,
cos²θ + (1/cosec²θ) + 17 = x
⇒ cos²θ + sin²θ + 17 = x   ; [1/cosecθ = sinθ]
⇒ 1 + 17 = x
⇒ x = 18
⇒ x² = 18²
⇒ x² = 324

∴ The value of x² is 324.

Question: The greatest value of sin⁴θ + cos⁴θ is?

Solution:
We know,
sin²θ + cos²θ = 1
⇒ (sin²θ + cos²θ)² = 1²    ; Squaring both sides
⇒ sin⁴θ + cos⁴θ + 2sin²θ.cos²θ = 1 
⇒ sin⁴θ + cos⁴θ = 1 - 2sin²θ.cos²θ
⇒ sin⁴θ + cos⁴θ = 1 - 0    ; [Put θ = 90° , cos90° = 0]
∴ sin⁴θ + cos⁴θ = 1 

So the greatest value of sin⁴θ + cos⁴θ is 1.

Question: If θ be an acute angle and 7sin²θ + 3cos²θ = 4, then the value of tanθ is?

Solution:
7sin²θ + 3cos²θ = 4
⇒ 7sin²θ + 3(1 - sin²θ) = 4
⇒ 7sin²θ + 3 - 3sin²θ = 4
⇒ 4sin²θ = 1
⇒ sin²θ = 1/4
⇒ sinθ = 1/2
⇒ sinθ = sin30°
∴ θ = 30°

∴ tanθ = tan30° = 1/√3

Question:  If 1 + sinθ = mcosθ than what is the value of cotθ? 

Solution:
Given that,
1+ sinθ = m cos θ
⇒ (1 + sinθ)/cosθ = m
⇒ (1/cosθ) + (sinθ/cosθ) = m
∴ secθ + tanθ = m ...............(i)

We know,
(secθ + tanθ) (secθ - tanθ) = 1
⇒ m(secθ - tanθ) = 1
⇒ secθ - tanθ = 1/m .................(ii)

Now, (i) - (ii) ⇒
secθ + tanθ - (secθ - tanθ) = m - (1/m)
⇒ secθ + tanθ - secθ + tanθ = (m² - 1)/m
⇒ 2tanθ = (m² - 1)/m
⇒  tanθ = (m² - 1)/2m
⇒  1/cotθ = 1/{(m² - 1)/2m}
∴ cotθ = 2m/(m² - 1)