Bank Math Master

Question: If sinθ/cosθ = 1, then find θ.

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

Question: If tanθ = 9/40, then secθ = ?

Solution:
এখানে,
tanθ = 9/40 = লম্ব/ভূমি

∴ লম্ব = 9, ভূমি = 40
∴ অতিভুজ = √(9² + 40²)
= √(81 + 1600)
= √1681 = 41

∴ secθ = 1/cosθ = অতিভুজ/ভূমি
= 41/40

Question: What is the maximum value of cos θ?

Solution:
cosθ এর সর্বনিম্ন মান −1 এবং সর্বোচ্চ মান 1
∴ cosθ এর সর্বোচ্চ মান = 1

Question: Which trigonometric ratio is undefined at 90°?

Solution:
sin90° = 1
cos90° = 0
cot90°= 0

∴ tan90° = sin90°/cos90°
= 1 / 0
= ∞ (Undefined)

Question: If A = 30°, then what is the value of (1 - tan²A)/(1 + tan²A)?

Solution:
Here, A = 30°

Now,
(1 - tan²A)/(1 + tan²A)
= {1 - (tan30°)²}/{1 + (tan30°)²}
= {1 - (1/√3)²}/{1 + (1/√3)²}
= (1 - 1/3)/(1 + 1/3)
= (2/3)/(4/3)
= 1/2

Question: If secA + tanA = 7/3, then what is the value of secA - tanA?

Solution:
দেয়া আছে,
secA + tanA = 7/3

আমরা জানি,
sec²A - tan²A = 1
⇒ (secA + tanA)(secA - tanA) = 1
⇒ 7/3 (secA - tanA) = 1
∴ secA - tanA = 3/7

Question: If θ = 30°, then sec²θ − tan²θ = ?

Solution:
Given, θ = 30°

Now,
sec²θ - tan²θ
= (sec30°)² - (tan30°)²
= (2/√3)² - (1/√3)²
= 4/3 - 1/3
= 3/3
= 1

Question: Find the value of cosec(- π/6)

Solution:
cosec(- π/6)
= - cosec(π/6)
= - 1/sin(π/6)
= - 1/sin30°
= - 1/(1/2)
= - 2

Question: If 1 - sinθ = n cosθ, then find the value of cotθ.

Solution:
Given,
1 - sinθ = n cosθ
⇒ (1 - sinθ)/cosθ = n
⇒ (1/cosθ) - (sinθ/cosθ) = n
∴ secθ - tanθ = n ...............(i)

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

Now, (ii) - (i) ⇒
(secθ + tanθ) - (secθ - tanθ) = 1/n - n
⇒ secθ + tanθ - secθ + tanθ = 1/n - n
⇒ 2tanθ = (1 - n²)/n
⇒ tanθ = (1 - n²)/2n
⇒ 1/cotθ = (1 - n²)/2n
∴ cotθ = 2n/(1 - n²)

Question: If r sinθ = 3, r cosθ = 4, then find the value of (4 tanθ + 1).

Solution:
r sinθ = 3
r cosθ = 4

Now,
(r sinθ)/(r cosθ) = 3/4
⇒ sinθ/cosθ = 3/4
⇒ tanθ = 3/4
⇒ 4 tanθ = 4 × 3/4 = 3
⇒ 4 tanθ + 1 = 3 + 1
∴ 4 tanθ + 1 = 4

Question: sin(A + 45°) = √2/2, find the value of A.

Solution:
sin(A + 45°) = √2/2
⇒ sin(A + 45°) = 1/√2
⇒ sin(A + 45°) = sin45°
⇒ A + 45° = 45°
⇒ A = 45° − 45°
∴ A = 0°

Question: The greatest value of sin⁴2θ + cos⁴2θ is?

Solution:
sin²2θ + cos²2θ = 1

(sin²2θ + cos²2θ)² = 1²
⇒ sin⁴2θ + cos⁴2θ + 2 sin²2θ cos²2θ = 1
⇒ sin⁴2θ + cos⁴2θ = 1 − 2 sin²2θ cos²2θ [2 sin²2θ cos²2θ = 0 (when θ = 0° or 90°)]
∴ sin⁴2θ + cos⁴2θ = 1 

∴ greatest value = 1

Question: If tan(θ - 30°) = √3, then find sinθ.

Solution:
Given,
tan(θ - 30°) = √3
⇒ tan(θ - 30°) = tan60°
⇒ θ - 30° = 60°
∴ θ = 90°

Now,
sinθ = sin90° = 1

Question: If tan 6A = √3, then find A.

Solution:
tan 6A = √3
⇒ tan 6A = tan60°
⇒ 6A = 60°
⇒ A = 60°/6
∴ A = 10°

Question: Find the value of 1 + {tan²θ/(1 + secθ)}.

Solution:
1 + {tan²θ / (1 + secθ)}
= 1 + {(sec²θ − 1)/(1 + secθ)}
= (1 + secθ + sec²θ − 1)/(1 + secθ)
= (secθ + sec²θ)/(1 + secθ)
= secθ

Question: If cosecθ - cotθ = 1/5, then find the value of sinθ + 5cosθ.

Solution:
cosecθ - cotθ = 1/5
⇒ (1/sinθ) - (cosθ/sinθ) = 1/5
⇒ (1 - cosθ)/sinθ = 1/5
⇒ sinθ = 5(1 - cosθ)
∴ sinθ = 5 - 5cosθ

∴ sinθ + 5cosθ = (5 - 5cosθ) + 5cosθ
= 5 - 5cosθ + 5cosθ
= 5

Question:

Solution:

Question: cot³30° - 2sin60° = ?

Solution:
Given that,
cot³30° - 2sin60°
= (√3)³ - 2(√3/2)
= 3√3 - √3
= 2√3

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: Find the value of sin²130° + cos²130°

Solution:
We know, sin²θ + cos²θ = 1

∴ sin²130° + cos²130° = 1