ABC is a triangle inscribed in a circle with centre O. Let \alpha=\angleBAC, where 45^\circ

Correct Answer is : cosβ=1−tan2α1+tan2α

\cos\beta= {1-\tan^{2}\alpha}/{1+\tan^{2}\alpha}

\cos\beta= {1+\tan^{2}\alpha}/{1-\tan^{2}\alpha}

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

UPSC NDA I 2018 Mathematics Solved Paper

Question : 42 of 120

Marks: +1, -0

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?

[NDA I 2018]

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