To verify the equation [tex]\(\frac{\sin B - \sin C}{\sin B + \sin C} = \cot \frac{B+C}{2} \tan \frac{B-C}{2}\)[/tex], we will simplify both sides of the equation step by step and show that they are indeed equal.
### Left Side: [tex]\(\frac{\sin B - \sin C}{\sin B + \sin C}\)[/tex]
Using the sum-to-product identities, we can simplify the left side:
[tex]\[
\sin B - \sin C = 2 \cos \left(\frac{B + C}{2}\right) \sin \left(\frac{B - C}{2}\right)
\][/tex]
[tex]\[
\sin B + \sin C = 2 \sin \left(\frac{B + C}{2}\right) \cos \left(\frac{B - C}{2}\right)
\][/tex]
Substitute these identities into the original expression:
[tex]\[
\frac{\sin B - \sin C}{\sin B + \sin C} = \frac{2 \cos \left( \frac{B+C}{2} \right) \sin \left( \frac{B-C}{2} \right)}{2 \sin \left( \frac{B+C}{2} \right) \cos \left( \frac{B-C}{2} \right)}
\][/tex]
The factor of 2 cancels out:
[tex]\[
\frac{\cos \left( \frac{B+C}{2} \right) \sin \left( \frac{B-C}{2} \right)}{\sin \left( \frac{B+C}{2} \right) \cos \left( \frac{B-C}{2} \right)}
\][/tex]
We can now rewrite the fraction:
[tex]\[
\frac{\cos \left( \frac{B+C}{2} \right)}{\sin \left( \frac{B+C}{2} \right)} \cdot \frac{\sin \left( \frac{B-C}{2} \right)}{\cos \left( \frac{B-C}{2} \right)}
\][/tex]
Recognize these as trigonometric functions:
[tex]\[
\frac{\cos \left( \frac{B+C}{2} \right)}{\sin \left( \frac{B+C}{2} \right)} = \cot \left( \frac{B+C}{2} \right)
\][/tex]
[tex]\[
\frac{\sin \left( \frac{B-C}{2} \right)}{\cos \left( \frac{B-C}{2} \right)} = \tan \left( \frac{B-C}{2} \right)
\][/tex]
So the expression simplifies to:
[tex]\[
\cot \left( \frac{B+C}{2} \right) \tan \left( \frac{B-C}{2} \right)
\][/tex]
### Right Side:
The right side of the equation is exactly:
[tex]\[
\cot \left( \frac{B+C}{2} \right) \tan \left( \frac{B-C}{2} \right)
\][/tex]
Hence, we have shown that:
[tex]\[
\frac{\sin B - \sin C}{\sin B + \sin C} = \cot \left( \frac{B+C}{2} \right) \tan \left( \frac{B-C}{2} \right)
\][/tex]
Thus, the equation is verified and both sides are equal.
