Question

$\ln \Delta ABC,a\sin (B-C)+b\sin (C-A)+c\sin (A-B)=$

• A. $0$
• B. $a+b+c$
• C. $a^2+b^2+c^2$
• D. $2(a^2+b^2+c^2)$

Answer 1

The correct option is A $0$

Given, $a\sin (B-C)+b\sin (C-A+C\sin (A-B)$

$=\sum a\sin (B-C)$

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

$=2R\sum \sin (B+C)\sin (B-C)$

$=2R\sum (\sin B\cos C+\cos B\sin C)(\sin B\cos C-\cos B\sin C)$

$=2R\sum {\sin }^{2}B{\cos }^{2}C-{\cos }^{2}B{\sin }^{2}C$

$=2R\sum {\sin }^{2}B(1-{\sin }^{2}C)-{\cos }^2(1-{\sin }^{2}B){\sin }^{2}C$

$=2R\sum {\sin }^{2}B-{\sin }^{2}B{\sin }^{2}C-{\sin }^{2}C+{\sin }^{2}B{\sin }^{2}C$

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

$=2R\sum {\sin }^{2}B-2R\sum {\sin }^{2}C$

$=0$.