Bank Math Master

Question: The value of sin²65° + sin²25° + cos²35° + cos²55° is?

Question: If tanθ + cotθ = 3, then tan²θ + cot²θ is?

Solution:
দেওয়া আছে,
tanθ + cotθ = 3
⇒ (tanθ + cotθ)² = 3²
⇒ tan²θ + cot²θ + 2tanθ . cotθ = 9
⇒ tan²θ + cot²θ = 9 - 2 [∵ tanθ . cotθ = 1]
∴ tan²θ + cot²θ = 7

Question: If sin(θ + 15°) = 3/√12 then 2sin²θ = ?

Solution:
sin(θ + 15°) = 3/√12
⇒ sin(θ + 15°) = 3/(2√3)
⇒ sin(θ + 15°) = √3/2
⇒ sin(θ + 15°) = sin60°
⇒ θ + 15° = 60°
⇒ θ = 45°

Now,
2sin²θ = 2(sin 45°)²
= 2 . (1/√2)²
= 2 . 1/2
= 1

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

Solution:
দেওয়া আছে,
rsinθ = 1 ......... (1)
rcosθ = √3 .............. (2)

(1) ÷ (2) হতে পাই
⇒ rsinθ/rcosθ =1/√3
⇒ tanθ = 1/√3
⇒ √3tanθ = 1

এখন, √3tanθ + 1 = 1 + 1
∴ √3tanθ + 1 = 2

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: 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°

অথবা,
2sinθ - 1 = 0
⇒ sinθ = 1/2
⇒ sinθ = ‍sin30°
∴ θ = 30°
যেহেতু, θ সূক্ষ্মকোণ,  θ = 30°

Question: If secθ + tanθ = x, then tanθ is - 

Solution:
দেওয়া আছে,
secθ + tanθ = x ................. (1)

আমরা জানি,
sec²θ - tan²θ = 1
বা, (secθ + tanθ)(secθ - tanθ) = 1
বা, x(secθ - tanθ) = 1
বা, secθ - tanθ = 1/x ................ (2)

(1) - (2) হতে পাই,
(secθ + tanθ) - (secθ - tanθ) = x - (1/x)
বা, 2tanθ = (x² - 1)/x
∴ 2tanθ =  (x² - 1)/x

Question: If θ = 30° then the value of cosθ - sin2θ is -

Solution:
দেওয়া আছে,
 θ = 30°

∴ cos30° - sin(2 ×30°)
= cos30° - sin60°
= (√3/2) - (√3/2)
= (√3 - √3)/2
= 0/2
= 0

Question: What is the solution of
 

Solution:

Question: If sinC = 3/4, then cosC = ?

Solution:

আমরা জানি,
sinx = লম্ব/অতিভুজ = 3/4
আমরা জানি,
ভূমি = √(অতিভুজ)² - (লম্ব)²
বা, ভূমি = √(4)² - (3)²
বা, ভূমি = √(16 - 9)
∴ ভূমি = √7

আবার,
cosx = ভূমি/অতিভুজ
= √7/4

Question: Which one is incorrect?

Solution:
আমরা জানি,
sin²A + cos²A = 1
⇒ sin²A = 1 - cos²A
⇒ cos²A = 1 - sin²A

আবার,
sec²A - tan²A = 1
⇒ sec²A = 1 + tan²A
⇒ tan²A = sec²A - 1

এবং
cosec²A - cot²A = 1
⇒ cosec²A = 1 + cot²A
⇒ cosec²A - 1 = cot²A

Question: Find the value of 
is -

Solution:
এখানে,




= 1/4

Question: If A = 45° then

Solution:
A = 45°

এখন 
(1 - tan²A)/(1 + tan²A)
= (1 - tan²45°)/(1 + tan²45°)
= (1 - 1²)/(1 + 1²)
= (1 - 1)/(1 + 1)
= 0/2
= 0

Question: ∠B is the right angle of a right angles triangle ABC. If tanA = 1, then 2sinACosA = ?

Solution:

দেওয়া আছে,
tanA = 1
ধরি,
বিপরীত বাহু = সন্নিহিত বাহু = a
অতিভুজ = √(a² + a²) = √2 a

∴ sinA = a/√2 a = 1/√2
cosA =  a/√2 a = 1/√2

∴ 2sinACosA = 2 × (1/√2) × (1/√2)
= 2 × 1/2
= 1

Question: If tanθ = √3^{- 1} then what is the value of θ?

Solution:
দেওয়া আছে,
tanθ = √3^{- 1}
⇒ tanθ = √(1/3)
⇒ tanθ = 1/√3
⇒ tanθ = tan30°
∴ θ = 30°

Question: Which one is correct?

Solution:
বিভিন্ন কোণের ত্রিকোনমিতিক অনুপাতের ক্ষেত্রে:
⇒ sin(- θ) = - sinθ
⇒ cos(- θ) = cosθ
⇒ tan(- θ) = - tanθ
⇒ cosec(- θ) = - cosecθ
⇒ sec(- θ) = secθ
⇒ cot( - θ) = - cotθ

Question: Find the value of cos60° sin30° + sin60° cos30° is?

Solution:
cos60° sin30° + sin60° cos30°
= (1/2) . (1/2) + (√3/2) . (√3/2)
= 1/4 + 3/4
= 4/4
= 1

Question:  
What is the value of cosA + secA is - 

Solution:
প্রদত্ত চিত্র হতে,
ΔABC সমকোণী ত্রিভুজ হতে,
অতিভূজ AC = 2 এবং ভূমি, BC = √3

আমরা জানি,
(লম্ব)² = (অতিভূজ)² - (লম্ব)²
বা, AC² = (2)² - (√3)²
বা, AC² = 4 - 3
∴ AC = 1

cosA = AB/AC = 1/2
secA = AC/AB = 2/1 = 2
∴ cosA + secA = 1/2 + 2 = 5/2