In triangle ABC , prove that `tan (B-C)/2 =[( b-c) /(b+c )] cot A/2` , ` tan (C-A)/2 = [(c-a)/(c+a)]cot B/2 `,` tan (A-B)/2 = [(a-b)/(a+b)]cot c/2`

Explanation

To prove the identities given in the question, we will use the sine rule and some trigonometric identities. Let's break down the proof step by step.

### Step 1: Use the Sine Rule
We know from the sine rule that:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]
where \( k \) is a constant.
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