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$\mathrm{ABC}$ is a triangle inscribed in a circle with centre $\mathrm{O}$. Let $\alpha=$ $\angle \mathrm{BAC}$, where $45^{\circ} < \alpha < 90^{\circ} .$ Let $\beta=\angle \mathrm{BOC}$. Which one of
the following is correct?
Options:
the following is correct?
Solution:
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Verified Answer
The correct answer is:
$\cos \beta=\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}$
$\angle \mathrm{BAC}=\alpha, \angle \mathrm{BOC}=\beta$
We know, from figure, $\beta=2 \alpha$
$\therefore \cos \beta=\cos 2 \alpha$
$=\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}$
We know, from figure, $\beta=2 \alpha$
$\therefore \cos \beta=\cos 2 \alpha$
$=\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}$
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