Respuesta :
Answer:
a.
cos θ = 2/5, tan θ = ?
- tan θ = sin θ / cos θ
- tan θ = √sin² θ / (2/5)
- tan θ= 5√(1 - cos²θ) / 2
- tan θ = 5√(1 - 4/25) / 2
- tan θ = 5√(21/25) / 2
- tan θ = √21 / 2
b.
cosec θ = 7/3, cos θ = ?
- cosec θ = 1/ sin θ
- cosec θ = 1/ √(1 - cos²θ)
- √(1 - cos²θ) = 1 / cosec θ
- 1 - cos²θ = (1 / (7/3))²
- cos²θ = 1 - 9 / 49
- cos²θ = 40/49
- cos θ = √40/49
- cos θ = 2√10/7
c.
cot θ = 4/3, sec θ = ?
- cot θ = cos θ / sin θ
- cos θ = cot θ * sin θ
- cos θ = 4/3 * √(1 - cos²θ)
- 9cos²θ = 16(1 - cos²θ)
- 25cos²θ = 16
- cos²θ = 16/25
- cos θ √16/25
- cos θ = 4/5
- sec θ = 1/ cos θ
- sec θ = 1/ (4/5)
- sec θ = 5/4
d.
tan θ = 3, cosec θ = ?
- sin²θ + cos²θ = 1
- 1 + cos²θ/sin²θ = 1/ sin²θ
- 1 + 1/tan²θ = cosec²θ
- 1 + 1/9 = cosec²θ
- cosec²θ = 10/9
- cosec θ = √(10/9)
- cosec θ = √10 / 3