Question

In any $\Delta ABC$, the given relation is$\frac{b-c}{b+c}=\frac{\tan \left(\frac{B-C}{2}\right)}{\tan \left(\frac{B+C}{2}\right)}$

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{b-c}{b+c}$

$=\frac{2R\sin B-2R\sin C}{2R\sin B+2R\sin C}$ using sine rule

$=\frac{\sin B-\sin C}{\sin B+\sin C}$

$=\frac{2\sin \left(\frac{B-C}{2}\right)\cos \left(\frac{B+C}{2}\right)}{2\sin \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)}$

$=\frac{\sin \left(\frac{B-C}{2}\right)\cos \left(\frac{B+C}{2}\right)}{\sin \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)}$

$=\frac{\frac{\sin \left(\frac{B-C}{2}\right)}{\cos \left(\frac{B-C}{2}\right)}}{\frac{\sin \left(\frac{B+C}{2}\right)}{\cos \left(\frac{B+C}{2}\right)}}$

$=\frac{\tan \left(\frac{B-C}{2}\right)}{\tan \left(\frac{B+C}{2}\right)}$