ABC is a triangle inscribed in a circle with centre O. Let `alpha = angle BAC,` where `45^@ lt alpha lt 90^@`. Let `beta = angle BOC.` Which one of the following is correct ?

Options

A. `cos beta = (1-tan^(2)alpha)/(1+tan^(2)alpha)`
B. `cos beta=(1+tan^(2)alpha)/(1-tan^(2)alpha)`
C. `cos beta=(2 tan alpha)/(1+tan^(2)alpha)`
D. `sin beta = 2 sin^(2) alpha`

Answer

A

Explanation

`angle BAC = alpha, angle BOC = beta`

We know, from figure, `beta = 2 alpha`
`therefore cos beta = cos 2 alpha`
`=(1-tan^(2)alpha)/(1+tan^(2)alpha)`.