Question

plz answer with format only plz fast​

Answer 1

Answer:

4) Option (a) => 87/25

5) Option (a) => 0

6) Option (a) => 1

7) Option (a) => 90°

Solution:

4)Here we go ↓

sin0°•sin1°•sin2°•....•sin88°•sin89°•sin90°

= sin0°(sin1°•sin2°•...•sin88°•sin89°•sin90°)

= 0•(sin1°•sin2°•...•sin88°•sin89°•sin90°)

= 0

Hence,

Required answer is 0

5) Here we go ↓

We have ;

=> 3sinθ = 4cosθ

=> sinθ / cosθ = 4/3

=> tanθ = 4/3 = p/b

Here,

p = 4 , b = 3

Now,

Using Pythagoras theorem , we have ;

=> h² = p² + b²

=> h² = 4² + 3²

=> h² = 16 + 9

=> h² = 25

=> h = √25

=> h = 5

Thus,

sinθ = p/h = 4/5

cosθ = b/h = 3/5

Thus,

4sin²θ - 3cos²θ + 2

= 4•(4/5)² - 3•(3/5)² + 2

= 4•(16/25) - 3•(9/25) + 2

= 64/25 - 27/25 + 2

= (64 - 27)/25 + 2

= 37/25 + 2

= (37 + 50)/25

= 87/25

Hence,

Required answer is 87/25

6) Here we go ↓

We have ;

=> cos9α = sinα

=> cos9α = cos(90° - α)

=> 9α = 90° - α

=> 9α + α = 90°

=> 10α = 90°

=> α = 90°/10

=> α = 9°

Thus,

tan5α = tan(5×9°) = tan45° = 1

Hence,

Required value of tan5α is 1 .

7) Here we go ↓

We have ;

=> 2sin²θ - cos²θ = 2

=> 2sin²θ - (1 - sin²θ) = 2

=> 2sin²θ - 1 + sin²θ = 2

=> 3sin²θ = 2 + 1

=> 3sin²θ = 3

=> sin²θ = 3/3

=> sin²θ = 1

=> sinθ = √1

=> sinθ = 1

=> sinθ = sin90°

=> θ = 90°

Hence,

The required value of θ is 90°