Question

In $\Delta ABC.\sum \frac{a^{2}sin(B-C)}{sinB+sinC}$

• A. $a+b$
• B. $b+c$
• C. $c+a$
• D. $0$

Answer 1

The correct option is D $0$

In $△ABC,a=2R\sin A=2R\sin ({180}^{\circ }-(B+C\left)\right)=2R\sin (B+C)$

$\sum \frac{a^2\sin (B-C)}{\sin B+\sin C}=2R\sum \frac{a\sin (B+C)\sin (B-C)}{\sin B+\sin C}$

$=2R\sum \frac{a({\sin }^{2}B-{\sin }^{2}C)}{\sin B+\sin C}$ $(\because \sin (A+B\left)\sin \right(A-B)={\sin }^{2}A-{\sin }^{2}B)$

$=4R^2\sum \sin A(\sin B-\sin C)$

$=4R^2\left(\sin A\right(\sin B-\sin C)+\sin B(\sin C-\sin A)+\sin C(\sin A-\sin B)$

$=0$