Question

In $△ABC,$ If $A={60}^{\circ }$ then $\frac{b}{c+a}+\frac{c}{a+b}=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$\begin{array}{c}B+C={120}^{\circ }\\ \frac{B}{2}+\frac{C}{2}={60}^{\circ }\\ \frac{B}{2}-{30}^{\circ }={30}^{\circ }-\frac{C}{2}\\ Now,\\ \frac{b}{c+a}+\frac{c}{a+b}-\frac{\sin B}{\sin A+\sin C}+\frac{\sin C}{\sin A+\sin B}\\ =\frac{\sin B}{2\sin \left(\frac{C+A}{2}\right)\cos \left(\frac{A-C}{2}\right)}+\frac{\sin C}{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}\\ =\frac{\sin \frac{B}{2}}{\cos \left(\frac{A-C}{2}\right)}+\frac{\sin \frac{C}{2}}{\cos \left(\frac{A-B}{2}\right)}\\ =\frac{\sin \frac{B}{2}}{\cos \left({30}^{\circ }-\frac{C}{2}\right)}+\frac{\sin \left(\frac{C}{2}\right)}{\cos \left({30}^{\circ }-\frac{B}{2}\right)}\\ =\frac{\sin \frac{B}{2}}{\cos \left(\frac{B}{2}-{30}^{\circ }\right)}+\frac{\sin \frac{C}{2}}{\cos \left({30}^{\circ }-\frac{B}{2}\right)}\\ =\frac{1}{\cos \left({30}^{\circ }-\frac{B}{2}\right)}2\sin {30}^{\circ }\cos \left(\frac{B-C}{4}\right)\\ =2\times \frac{1}{2}\times \frac{1}{\cos \left({30}^{\circ }-\frac{B}{2}\right)}\times \cos \left(\frac{B}{2}-{30}^{\circ }\right)\\ =1\\ hence,optionAisthecorrectanswer.\end{array}$