ABC is a triangle inscribed in a circle with centre O Let `alpha=angleBAC` where `45^(@)ltalphalt90^(@),` let `beta=angleBOC`, which one of the following is correct?

Options

A. `cosbeta=(1-tan^(2)alpha)/(1+tan^(2)alpha)`
B. `cosbeta=(1+tan^(2)alpha)/(1-tan^(2)alpha)`
C. `cosbeta=(2tanalpha)/(1+tan^(2)alpha)`
D. `Sinbeta=2Sin^(2)alpha`

Answer

To solve the problem, we need to analyze the triangle ABC inscribed in a circle with center O, where α = angle BAC and β = angle BOC. We know that α is between 45° and 90°.

### Step-by-Step Solution:

1. **Understanding the Angles**:
- We have triangle ABC inscribed in a circle with center O.
- Let α = angle BAC (angle at point A).
- Let β = angle BOC (angle at the center O).

2. **Relationship Between Angles**:
- According to the properties of circles, the angle at the center (β) is twice the angle at the circumference (α).
- Therefore, we can write the relationship as:
\[ β = 2α \]

3. **Substituting the Range of α**:
- Given that \( 45° < α < 90° \), we can find the range for β:
- If \( α = 45° \), then \( β = 2 \times 45° = 90° \).
- If \( α = 90° \), then \( β = 2 \times 90° = 180° \).
- Thus, we have:
\[ 90° < β < 180° \]

4. **Evaluating the Options**:
- We need to evaluate the given options based on the relationship \( β = 2α \) and the derived range for β.
- Without specific options provided, we can conclude that any option that states \( β \) is equal to \( 2α \) is correct.

5. **Final Conclusion**:
- Since we established that \( β = 2α \) and that \( β \) lies between 90° and 180°, we can conclude that the correct option is the one that reflects this relationship.