Answer:
1. Cosθ / SineθCosθ
2. Sineθ / Cos²θSineθ
3. Cosθ / Sineθ
4. Cos²θ + Sin²θ – Sin²θ
Step-by-step explanation:
1. SecθCotθ
Recall
Sec θ = 1/Cos θ
Cot θ = 1/Tan θ
But Tan θ = Sine θ / Cos θ
Thus,
Cot θ = 1 ÷ Sine θ / Cos θ
Cot θ = 1 × Cos θ / Sine θ
Cot θ = Cos θ / Sine θ
Therefore,
SecθCotθ = 1/Cos θ × Cos θ / Sine θ
SecθCotθ = Cosθ / SineθCosθ
2. SecθTanθCscθ
Recall
Sec θ = 1/Cos θ
Tan θ = Sine θ / Cos θ
Csc θ = 1/Sine θ
Thus,
SecθTanθCscθ =
1/Cosθ × Sineθ/Cosθ × 1/Sineθ
= Sineθ / Cos²θSineθ
3. Cscθ/Secθ
Recall
Csc θ = 1/Sine θ
Sec θ = 1/Cos θ
Thus,
Cscθ/Secθ = 1/Sine θ ÷ 1/Cos θ
= 1/Sine θ × Cos θ
= Cosθ / Sineθ
4. Cosθ / Secθ
Recall
Sec θ = 1/Cos θ
Cosθ / Secθ = Cosθ ÷ 1/Cosθ
= Cosθ × Cosθ
= Cos²θ
Recall
Cos²θ + Sin²θ = 1
Cos²θ = 1 – Sin²θ
But
1 = Cos²θ + Sin²θ
Thus,
Cos²θ = Cos²θ + Sin²θ – Sin²θ
Therefore,
Cosθ / Secθ = Cos²θ + Sin²θ – Sin²θ
Answer:
a. 11.26 % b. 6.76 %. It appears so since 6.76 % ≠ 15 %
Step-by-step explanation:
a. This is a binomial probability.
Let q = probability of giving out wrong number = 15 % = 0.15
p = probability of not giving out wrong number = 1 - q = 1 - 0.15 = 0.75
For a binomial probability, P(x) = ⁿCₓqˣpⁿ⁻ˣ. With n = 10 and x = 1, the probability of getting a number wrong P(x = 1) = ¹⁰C₁q¹p¹⁰⁻¹
= 10(0.15)(0.75)⁹
= 1.5(0.0751)
= 0.1126
= 11.26 %
b. At most one wrong is P(x ≤ 1) = P(0) + P(1)
= ¹⁰C₀q⁰p¹⁰⁻⁰ + ¹⁰C₁q¹p¹⁰⁻¹
= 1 × 1 × (0.75)¹⁰ + 10(0.15)(0.75)⁹
= 0.0563 + 0.01126
= 0.06756
= 6.756 %
≅ 6.76 %
Since the probability of at most one wrong number i got P(x ≤ 1) = 6.76 % ≠ 15 % the original probability of at most one are not equal, it thus appears that the original probability of 15 % is wrong.
