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Question
If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`
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Solution
In ∆ABC, we have
A + B + C = 180º
⇒ B + C = 180º – A
`\Rightarrow \frac{B+C}{2}=\text{ }90^\text{o}-\frac{A}{2}`
Taking tan on both sides, we get
`\Rightarrow \tan ( \frac{B+C}{2})=\tan( 90^\text{o}-\frac{A}{2})`
`\Rightarrow \tan ( \frac{B+C}{2} )=\cot \frac{A}{2}`
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