ABC is a triangle inscribed in a circle with centre O. Let α = ∠BAC, where 45° < α < 90°. Let β = ∠BOC. Which one of the following is correct?

\(\cos {\rm{\beta }} = \frac{{1 + {{\tan }^2}{\rm{\alpha }}}}{{1 - {{\tan }^2}{\rm{\alpha }}}}\)

\(\cos {\rm{\beta }} = \frac{{1 - {{\tan }^2}{\rm{\alpha }}}}{{1 + {{\tan }^2}{\rm{\alpha }}}}\)

\(\cos {\rm{\beta }} = \frac{{2\tan {\rm{\alpha }}}}{{1 + {{\tan }^2}{\rm{\alpha }}}}\)

ABC is a triangle inscribed in a circle with centre O. Let α = ∠BAC

1.	ABC is a triangle inscribed in a circle with centre O. Let α = ∠BAC, where 45° < α < 90°. Let β = ∠BOC. Which one of the following is correct?
A.	\(\cos {\rm{\beta }} = \frac{{1 - {{\tan }^2}{\rm{\alpha }}}}{{1 + {{\tan }^2}{\rm{\alpha }}}}\)
B.	\(\cos {\rm{\beta }} = \frac{{1 + {{\tan }^2}{\rm{\alpha }}}}{{1 - {{\tan }^2}{\rm{\alpha }}}}\)
C.	\(\cos {\rm{\beta }} = \frac{{2\tan {\rm{\alpha }}}}{{1 + {{\tan }^2}{\rm{\alpha }}}}\)
D.	sin β = 2 sin2 α
Answer» B. \(\cos {\rm{\beta }} = \frac{{1 + {{\tan }^2}{\rm{\alpha }}}}{{1 - {{\tan }^2}{\rm{\alpha }}}}\)