A) \[abc\]
B) \[3abc\]
C) \[a+b+c\]
D) None of these
Correct Answer: B
Solution :
\[{{a}^{3}}\cos (B-C)+{{b}^{3}}\cos (C-A)+{{c}^{3}}\cos (A-B)\] \[={{k}^{3}}{{\sin }^{3}}A\cos (B-C)+{{k}^{3}}{{\sin }^{3}}B\cos (C-A)\]\[+{{k}^{3}}{{\sin }^{3}}C\cos (A-B)\] \[=\frac{1}{2}{{k}^{3}}[{{\sin }^{2}}A(\sin 2B+\sin 2C)+{{\sin }^{2}}B(\sin 2C+\sin 2A)\]\[+{{\sin }^{2}}C(\sin 2A+\sin 2B)]\] \[={{k}^{3}}[\sin A\sin B(\sin A\cos B+\cos A\sin B)\] \[+\sin B\sin C(\sin B\cos C+\cos B\sin C)\] \[+\sin C\sin A(\sin C\cos A+\cos C\sin A)]\] \[={{k}^{3}}[\sin A\sin B\sin C+\sin B\sin C\sin A+\sin C\sin A\sin B]\] \[=3{{k}^{3}}\sin \,A\,\sin \,B\,\,C=3abc\]
You need to login to perform this action.
You will be redirected in
3 sec
