Multiple choice

$ABC$ is a triangle inscribed in a circle with centre $O$. Let $\alpha =\angle BAC$, where ${ 45 }^{ o }<\alpha <{ 90 }^{ o }$. Let $ \beta =\angle BOC$. Which one of the following is correct?

  1. $\cos { \beta } =\cfrac { 1-\tan ^{ 2 }{ \alpha } }{ 1+\tan ^{ 2 }{ \alpha } } $
  2. $\cos { \beta } =\cfrac { 1+\tan ^{ 2 }{ \alpha } }{ 1-\tan ^{ 2 }{ \alpha } } $
  3. $\cos { \beta } =\cfrac { 2\tan { \alpha } }{ 1+\tan ^{ 2 }{ \alpha } } $
  4. $\sin { \beta } =2\sin ^{ 2 }{ \alpha } $
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A Correct answer
Explanation

In a circle, the angle at the center (beta) is twice the angle at the circumference (alpha) subtended by the same arc, so beta = 2 * alpha. Thus, cos(beta) = cos(2 * alpha). Using the double angle identity, cos(2 * alpha) = (1 - tan^2(alpha)) / (1 + tan^2(alpha)).

AI explanation

By the Inscribed Angle Theorem, the central angle beta is twice the inscribed angle alpha, meaning beta equals 2 alpha. Using the double angle formula for cosine, we get cos of beta equals cos of 2 alpha, which simplifies to the expression 1 minus tan squared alpha over 1 plus tan squared alpha. The result is cos beta equals the fraction 1 minus tan squared alpha over 1 plus tan squared alpha.

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