@Ashley Shamidha
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Then expand to get
\(\tan A (\tan B+\tan C) + \tan B \tan C\).

It's easy to prove \(\frac{\sin A} {\cos B \cos C}=\tan B + \tan C\)

Then after some simplification,you will get the required expression as,
\(1+\sec A \sec B \sec C\), which is always greater than 2 in case of triangle.

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## Comments

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TopNewestPlease someone prove this

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What are the the restrictions on B and C?

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ABC is a triangle and A+B+C =180

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It's easy to prove \(\frac{\sin A} {\cos B \cos C}=\tan B + \tan C\)

Then after some simplification,you will get the required expression as, \(1+\sec A \sec B \sec C\), which is always greater than 2 in case of triangle.

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