Prove the following identities:

1. [tex]\[ \frac{\cos B + \cos C}{\sin B - \sin C} = \cot \frac{B - C}{2} \][/tex]

2. [tex]\[ \frac{\cos B - \cos C}{\sin B + \sin C} = -\tan \frac{B - C}{2} \][/tex]



Answer :

Let's prove the given trigonometric identities in a step-by-step manner.

### Proving the First Identity
[tex]\[ \frac{\cos B + \cos C}{\sin B - \sin C} = \cot \frac{B - C}{2} \][/tex]

To prove this, we use sum-to-product identities.

1. Sum-to-Product for Cosine:
[tex]\[ \cos B + \cos C = 2 \cos \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right) \][/tex]

2. Sum-to-Product for Sine:
[tex]\[ \sin B - \sin C = 2 \cos \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right) \][/tex]

Now, substituting these identities into our original fraction:

[tex]\[ \frac{\cos B + \cos C}{\sin B - \sin C} = \frac{2 \cos \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)}{2 \cos \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)} \][/tex]

3. Simplify the expression by canceling common factors:

[tex]\[ \frac{2 \cos \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)}{2 \cos \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)} = \frac{\cos \left( \frac{B - C}{2} \right)}{\sin \left( \frac{B - C}{2} \right)} \][/tex]

4. Recognize the quotient as cotangent:

[tex]\[ \frac{\cos \left( \frac{B - C}{2} \right)}{\sin \left( \frac{B - C}{2} \right)} = \cot \left( \frac{B - C}{2} \right) \][/tex]

Thus, we have proved the first identity:

[tex]\[ \frac{\cos B + \cos C}{\sin B - \sin C} = \cot \frac{B - C}{2} \][/tex]

### Proving the Second Identity
[tex]\[ \frac{\cos B - \cos C}{\sin B + \sin C} = -\tan \frac{B - C}{2} \][/tex]

Similarly, we use sum-to-product identities for the second proof.

1. Sum-to-Product for Cosine:
[tex]\[ \cos B - \cos C = -2 \sin \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right) \][/tex]

2. Sum-to-Product for Sine:
[tex]\[ \sin B + \sin C = 2 \sin \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right) \][/tex]

Now, substituting these identities into our original fraction:

[tex]\[ \frac{\cos B - \cos C}{\sin B + \sin C} = \frac{-2 \sin \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)}{2 \sin \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)} \][/tex]

3. Simplify the expression by canceling common factors:

[tex]\[ \frac{-2 \sin \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)}{2 \sin \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)} = \frac{- \sin \left( \frac{B - C}{2} \right)}{\cos \left( \frac{B - C}{2} \right)} \][/tex]

4. Recognize the quotient as tangent:

[tex]\[ \frac{- \sin \left( \frac{B - C}{2} \right)}{\cos \left( \frac{B - C}{2} \right)} = -\tan \left( \frac{B - C}{2} \right) \][/tex]

Thus, we have proved the second identity:

[tex]\[ \frac{\cos B - \cos C}{\sin B + \sin C} = -\tan \frac{B - C}{2} \][/tex]

Both identities are now proven.