Question

If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`

Answer 1

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}`