Question

$ABC$ 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?

• A. $\cos \beta =\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$
• B. $\cos \beta =\frac{1+{\tan }^2\alpha }{1-{\tan }^2\alpha }$
• C. $\cos \beta =\frac{2\tan \alpha }{1+{\tan }^2\alpha }$
• D. $\sin \beta =2{\sin }^2\alpha$

Answer 1

The correct option is A $\cos \beta =\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$

As we know that, angle at center is twice, the angle at circumference

Therefore, $\beta =2\alpha$

and $cos\beta =cos2\alpha =co{s}^2\alpha -si{n}^2\alpha =\frac{1-ta{n}^2\alpha }{1+ta{n}^2\alpha }$

Therefore, Answer is $A$